There are 10 professors and 20 lecturers, out of whom a committee of 2 professors and 3 lecturers is to be formed. Find: C1 | C2 (a) In how many ways can the committee be formed: | (i) 10C2 × 19C3 (b) In how many ways a particular professor is included: | (ii) 10C2 × 19C2 (c) In how many ways a particular lecturer is included: | (iii) 9C1 × 20C3 (d) In how many ways a particular lecturer is excluded: | (iv) 10C2 × 20C3

This is a matching answer type question as classified in NCERT Exemplar ( a ) W e h a v e t o s e l e c t 2 p r o f e s s o r o u t o f 1 0 a n d 3 l e c t u r e r s o u t o f 2 0 ∴ N u m b e r o f w a y s o f s e l e c t i o n = C 2 1 0 × C 3 2 0 ( b ) W h e n a p a r t i c u l a r p r o f e s s o r i s i n c l u d e d t a k e n n u m b e r o f w a y s = C 1 1 0 − 1 × C 3 2 0 = C 1 9 × C 3 2 0 ( c ) W h e n a p a r t i c u l a r l e c t u r e r i s i n c l u d e d , t h e n n u m b e r o f w a y s = C 2 1 0 × C 2 1 9 ( d ) W h e n a p a r t i c u l a r l e c t u r e r i s e x c l u d e d , t h e n n u m b e r o f w a y s = C 2 1 0 × C 3 1 9 H e n c e , t h e r e q u i r e d m a t c h i n g i s ( a ) ↔ ( i v ) , ( b ) ↔ ( i i i ) , ( c ) ↔ ( i i ) a n d ( d ) ↔ ( i )

How many words (with or without dictionary meaning) can be made from the letters of the word MONDAY, assuming that no letter is repeated, if C1 | C2 (a) 4 letters are used at a time | (i) 720 (b) All letters are used at a time | (ii) 240 (c) All letters are used but the first is a vowel | (iii) 360

This is a matching answer type question as classified in NCERT Exemplar ( a ) 4 l e t t e r s a r e u s e d a t a t i m e = P 4 6 = 6 ! 2 ! = 3 6 0 ( b ) A l l l e t t e r s a r e u s e d a t a t i m e = P 6 6 = 6 ! = 7 2 0 ( c ) A l l l e t t e r s a r e u s e d b u t f i r s t l e t t e r i s v o w e l = 2 × 5 ! = 2 × 1 2 0 = 2 4 0 H e n c e , t h e r e q u i r e d m a t c h i n g i s ( a ) ↔ ( i i i ) , ( b ) ↔ ( i ) , ( c ) ↔ ( i i )

There are 3 books on Mathematics, 4 on Physics, and 5 on English. How many different collections can be made such that each collection consists of: C1 | C2 (a) One book of each subject; | (i) 3968 (b) At least one book of each subject: | (ii) 60 (c) At least one book of English: | (iii) 3255

This is a matching answer type question as classified in NCERT Exemplar W e h a v e 3 b o o k s o f M a t h e m a t i c s , 4 o f P h y s i c s a n d 5 o n E n g l i s h ( a ) O n e b o o k o f e a c h s u b j e c t = C 1 3 × C 1 4 × C 1 5 = 3 × 4 × 5 = 6 0 ( b ) A t l e a s t o n e b o o k o f e a c h s u b j e c t = ( 2 3 − 1 ) × ( 2 4 − 1 ) × ( 2 5 − 1 ) = 7 × 1 5 × 3 1 = 3 2 5 5 ( c ) A t l e a s t o n e b o o k o f E n g l i s h = ( 2 5 − 1 ) × 2 7 = 3 1 × 1 2 8 = 3 9 6 8 H e n c e , t h e r e q u i r e d m a t c h i n g i s ( a ) ↔ ( i i ) , ( b ) ↔ ( i i i ) a n d ( c ) ↔ ( i )

If npr = 840, nCr = 35, then r = ______.

This is an Objective Type Questions as classified in NCERT Exemplar G i v e n t h a t P r n = 8 4 0 a n d C r n = 3 5 ⇒ n ! ( n − r ) ! = 8 4 0 … ( i ) a n d n ! r ! ( n − r ) ! = 3 5 … ( i i ) D i v i d i n g e q n . ( i ) b y ( i i ) w e g e t n ! ( n − r ) ! n ! r ! ( n − r ) ! = 8 4 0 3 5 ⇒ n ! ( n − r ) ! × r ! ( n − r ) ! n ! = 2 4 ⇒ r ! = 2 4 ⇒ r ! = 4 × 3 × 2 × 1 ⇒ r ! = 4 ! ∴ r = 4 H e n c e , t h e v a l u e o f t h e f i l l e r i s 4 .

Eight chairs are numbered 1 to 8. Two women and 3 men wish to occupy one chair each. First the women choose the chairs from amongst the chairs 1 to 4 and then men select from the remaining chairs. Find the total number of possible arrangements. [Hint: 2 women occupy the chair, from 1 to 4 in 4P2 ways and 3 men occupy the remaining chairs in 6P3 ways.]

This is a short answer type question as classified in NCERT Exemplar W e h a v e 2 w o m e n a n d 3 m e n F i r s t w o m e n c h o o s e t h e c h a i r s a m o n g s t t h e c h a i r s 1 t o 4 i . e . T o t a l n u m b e r o f c h a i r s = 4 S o , t h e n u m b e r o f a r r a n g e m e n t s = P 2 4 w a y s N o w 3 m e n c h o o s e f r o m t h e r e m a i n i n g 6 c h a i r s S o , t h e n u m b e r o f a r r a n g e m e n t s = P 3 6 w a y s ∴ T o t a l n u m b e r o f a r r a n g e m e n t s = P 2 4 × P 3 6 = 4 ! ( 4 − 2 ) ! × 6 ! ( 6 − 3 ) ! = 4 ! 2 ! × 6 ! 3 ! = 4 . 3 . 2 ! 2 ! × 6 . 5 . 4 . 3 ! 3 ! = 1 2 × 1 2 0 = 1 4 4 0 H e n c e , t h e t o t a l n u m b e r o f p o s s i b l e a r r a n g e m e n t s a r e 1 4 4 0 .

If the letters of the word RACHIT are arranged in all possible ways as listed in dictionary, then what is the rank of the word RACHIT? [Hint: In each case number of words beginning with A, C, H, I is 5!]

This is a short answer type question as classified in NCERT Exemplar T h e a l p h a b e t i c a l o r d e r o f R A C H I T i s A , C , H , I , R a n d T N u m b e r o f w o r d s b e g i n n i n g w i t h A = 5 ! N u m b e r o f w o r d s b e g i n n i n g w i t h C = 5 ! N u m b e r o f w o r d s b e g i n n i n g w i t h H = 5 ! N u m b e r o f w o r d s b e g i n n i n g w i t h I = 5 ! a n d N u m b e r o f w o r d s b e g i n n i n g w i t h R i . e . R A C H I T = 1 ∴ T h e r a n k o f t h e w o r d ' R A C H I T ' i n t h e d i c t i o n a r y = 5 ! + 5 ! + 5 ! + 5 ! + 1 = 4 × 5 ! + 1 = 4 × 5 . 4 . 3 . 2 . 1 + 1 = 4 × 1 2 0 + 1 = 4 8 0 + 1 = 4 8 1

Out of 18 points in a plane, no three are in the same line except five points which are collinear. Find the number of lines that can be formed joining the points. [Hint: Number of straight lines = 18C2 – 5C2 + 1.]

This is a long answer type question as classified in NCERT Exemplar Total number of points=12 Out of 18 numbers, 5 are collinear and we get a straight line by joining any two points. ∴ Total number of straight line formed by joining 2 points out of 18 points=C218 Number of straight lines formed by joining 2 points out of 5 points=C25 But 5 points are collinear and we get only one line when they are joined pairwise. S o , t h e r e q u i r e d n u m b e r o f s t r a i g h t l i n e s a r e = C 2 1 8 − C 2 5 + 1 = 1 8 . 1 7 2 . 1 − 5 . 4 2 . 1 + 1 = 1 5 3 − 1 0 + 1 = 1 4 4 H e n c e , t o t a l n u m b e r o f s t r a i g h t l i n e s = 1 4 4 .

We wish to select 6 persons from 8, but if the person A is chosen, then B must be chosen. In how many ways can selections be made?

This is a long answer type question as classified in NCERT Exemplar T o t a l n u m b e r o f p e r s o n s = 8 N u m b e r o f p e r s o n s t o b e s e l e c t e d = 6 C o n d i t i o n i s t h a t i f A i s c h o o s e n , B m u s t b e c h o o s e n C a s e I : W h e n A i s c h o o s e n , B m u s t b e c h o o s e n N u m b e r o f w a y s = C 4 6 [ ? A a n d B a r e s e t t o b e c h o o s e n ] C a s e I I : W h e n A i s n o t c h o o s e n , t h e n B m a y b e c h o o s e n N u m b e r o f w a y s = C 6 7 S o , t h e t o t a l n u m b e r o f w a y s = C 4 6 + C 6 7 [ ? T h e r e a r e t w o c a s e s ] = C 2 6 + C 1 7 [ ? C r n = C n − r n ] = 6 . 5 2 . 1 + 7 = 1 5 + 7 = 2 2 w a y s H e n c e , t h e r e q u i r e d n u m b e r o f w a y s = 2 2 .

If nC12 = nC8, then n is equal to: (a) 20 (b) 12 (c) 6 (d) 30

This is an Objective Type Questions as classified in NCERT Exemplar G i v e n t h a t : C 1 2 n = C 8 n ⇒ C n − 1 2 n = C 8 n [ ? C r n = C n − r n ] ⇒ n − 1 2 = 8 ⇒ n = 8 + 1 2 = 2 0 H e n c e , t h e c o r r e c t o p t i o n i s ( a ) .

If y = y(x) is the solution of the differential equation, dy/dx + 2ytanx = sinx, y(π/3) = 0, then the maximum value of the function y(x) over R is equal to :

1/2

The range of a ∈ R for which the function f(x) = (4a - 3)(x + log?5) + 2(a - 7)cot(x/2)sin²(x/2), x ≠ 2nπ, n ∈ N has critical points is:

(-∞, -1]

(-3, 1)

Let a vector αi + βj be obtained by rotating the vector √3i + j by an angle 45° about the origin in counterclockwise direction in the first quadrant. Then the area of triangle having vertices (α, β), (0, β) and (0, 0) is equal to :

1/2

2√2

If for a > 0, the feet of perpendiculars from the points A(a, -2a, 3) and B(0, 4, 5) on the plane lx + my + nz = 0 are points C(0, -a, -1) and D respectively, then the length of line segment CD is equal to :

√66

√41

If the three normals drawn to the parabola, y² = 2x pass through the point (a, 0) a ≠ 0, then 'a' must be greater than :

1/2

The locus of the midpoints of the chord of the circle, x² + y² = 25 which is tangent to the hyperbola, x²/9 - y²/16 = 1 is:

(x² + y²)² - 9x² + 16y² = 0

(x² + y²)² - 9x² + 144y² = 0

(x² + y²)² - 9x² - 16y² = 0

(x² + y²)² - 16x² + 9y² = 0

Let A = [[-i, 1], [-i, i]], i = √-1. Then, the system of linear equation A?[x; y] = [8; 64] has :

Infinitely many solutions

Exactly two solutions

Let S? = ∑(r=1 to k) tan?¹(6? / (2²??¹ + 3²??¹)). Then lim (k→∞) S? is equal to :

tan⁻¹(3/2)

tan⁻¹(3)

If for x ∈ (0, π/2), log??sinx + log??cosx = -1 and log??(sinx + cosx) = ½(log??n - 1), n > 0, then the value of n is equal to :

16

Let the position vectors of two points P and Q be 3i - j + 2k and i + 2j - 4k respectively. Let R and S be two points such that the direction rations of lines PR and QS are (4, -1, 2) and (-2, 1, -2), respectively. Let lines PR and QS intersect at T. If the vector TA is perpendicular to both PR and QS and the length of vector TA is √5 units, then the modulus of a position vector of A is :

√482

Which of the following Boolean expression is a tautology?

(p ∨ q) ∨ (p → q)

Maths NCERT Exemplar Solutions Class 11th Chapter Seven: Overview, Questions, Preparation

Updated on Jul 3, 2025 10:23 IST

By Raj Pandey

Table of content

Permutations and Combinations Short Answer Type Questions
Permutations and Combinations Long Answer Type Questions
Permutations and Combinations Objective Type Questions
Permutations and Combinations Fill in the Blanks Type Questions
Permutations and Combinations True or False Type Questions
Permutations and Combinations Matching Type Questions

Permutations and Combinations Long Answer Type Questions

1. 18 mice were placed in two experimental groups and one control group, with all groups equally large. In how many ways can the mice be placed into three groups?

Sol.

\begin{array}{l} G i v e n t h a t 18 m i c e w e r e p l a c e d e q u a l l y i n t w o experimental g r o u p s a n d \\ o n e c o n t r o l g r o u p i . e . 3 g r o u p s \\ ∴ T h e r e q u i r e d n u m b e r o f a r r a n g e m e n t s = \frac{T o t a l a r r a n g e m e n t s}{E q u a l l y l i k e l y a r r a n g e m e n t s} \\ = \frac{18!}{6! 6! 6!} = \frac{18!}{{(6!)}^{3}} \\ H e n c e, t h e r e q u i r e d a r r a n g e m e n t s = \frac{18!}{{(6!)}^{3}} \end{array}

2. A bag contains six white marbles and five red marbles. Find the number of ways in which four marbles can be drawn from the bag if:

(i) They can be of any colour.

(ii) Two must be white and two red.

(iii) They must all be of the same colour.

Sol.

\begin{array}{l} T o t a l n u m b e r o f m a r b l e s = 6 w h i t e + 5 r e d = 11 m a r b l e s \\ (i) Since, w e h a v e t o d r a w 4 m a r b l e s o f a n y c o l o u r f r o m t h e 11 m a r b l e s \\ ∴ Required n u m b e r o f w a y s = {}_{11}{C_{4}} \\ (i i) I f 2 m u s t b e w h i t e a n d 2 m u s t b e r e d, t h e n t h e r e q u i r e d n u m b e r o f w a y s = {}_{6}{C_{2}} \times {}_{5}{C_{2}} \\ (i i i) I f a l l t h e 4 m a r b l e s a r e o f t h e s a m e c o l o u r, t h e n t h e r e q u i r e d n u m b e r o f w a y s = {}_{6}{C_{4}} + {}_{5}{C_{4}} \\ H e n c e, t h e r e q u i r e d n u m b e r o f w a y s a r e \\ (i) {}_{11}{C_{4}} (i i) {}_{6}{C_{2}} \times {}_{5}{C_{2}} (i i i) {}_{6}{C_{4}} + {}_{5}{C_{4}} \end{array}

Commonly asked questions

18 mice were placed in two experimental groups and one control group, with all groups equally large. In how many ways can the mice be placed into three groups?

This is a long answer type question as classified in NCERT Exemplar

$\begin{array}{l} G i v e n t h a t 18 m i c e w e r e p l a c e d e q u a l l y i n t w o experimental g r o u p s a n d \\ o n e c o n t r o l g r o u p i . e . 3 g r o u p s \\ ∴ T h e r e q u i r e d n u m b e r o f a r r a n g e m e n t s = \frac{T o t a l a r r a n g e m e n t s}{E q u a l l y l i k e l y a r r a n g e m e n t s} \\ = \frac{18!}{6! 6! 6!} = \frac{18!}{{(6!)}^{3}} \\ H e n c e, t h e r e q u i r e d a r r a n g e m e n t s = \frac{18!}{{(6!)}^{3}} \end{array}$

A bag contains six white marbles and five red marbles. Find the number of ways in which four marbles can be drawn from the bag if:

(i) They can be of any colour.

(ii) Two must be white and two red.

(iii) They must all be of the same colour.

This is a long answer type question as classified in NCERT Exemplar

$\begin{array}{l} T o t a l n u m b e r o f m a r b l e s = 6 w h i t e + 5 r e d = 11 m a r b l e s \\ (i) Since, w e h a v e t o d r a w 4 m a r b l e s o f a n y c o l o u r f r o m t h e 11 m a r b l e s \\ ∴ Required n u m b e r o f w a y s = {}_{11}{C_{4}} \\ (i i) I f 2 m u s t b e w h i t e a n d 2 m u s t b e r e d, t h e n t h e r e q u i r e d n u m b e r o f w a y s = {}_{6}{C_{2}} \times {}_{5}{C_{2}} \\ (i i i) I f a l l t h e 4 m a r b l e s a r e o f t h e s a m e c o l o u r, t h e n t h e r e q u i r e d n u m b e r o f w a y s = {}_{6}{C_{4}} + {}_{5}{C_{4}} \\ H e n c e, t h e r e q u i r e d n u m b e r o f w a y s a r e \\ (i) {}_{11}{C_{4}} (i i) {}_{6}{C_{2}} \times {}_{5}{C_{2}} (i i i) {}_{6}{C_{4}} + {}_{5}{C_{4}} \end{array}$

In how many ways can a football team of 11 players be selected from 16 players? How many of them will:

(i) Include 2 particular players?

(ii) Exclude 2 particular players?

This is a long answer type question as classified in NCERT Exemplar

$\begin{array}{l} G i v e n t h a t t h e t o t a l n u m b e r o f p l a y e r s = 16 \\ W e h a v e t o s e l e c t 11 p l a y e r s o u t o f 16 p l a y e r s . \\ (i) I f 2 p l a y e r s a r e i n c l u d e d, t h e n n u m b e r o f w a y s o f s e l e c t i o n = {}_{16 - 2}{C_{11 - 2}} = {}_{14}{C_{9}} \\ (i i) I f 2 p l a y e r s a r e e x c l u d e d, t h e n n u m b e r o f w a y s o f s e l e c t i o n = {}_{16 - 2}{C_{11}} = {}_{14}{C_{11}} \\ H e n c e, t h e r e q u i r e d n u m b e r o f w a y s o f s e l e c t i o n a r e \\ (i) {}_{14}{C_{9}} (i i) {}_{14}{C_{11}} \end{array}$

A sports team of 11 students is to be constituted, choosing at least 5 from Class XI and at least 5 from Class XII. If there are 20 students in each of these classes, in how many ways can the team be constituted?

This is a long answer type question as classified in NCERT Exemplar

$\begin{array}{l} T o t a l n u m b e r o f s t u d e n t s i n e a c h c l a s s = 20 \\ W e h a v e t o s e l e c t a t l e a s t 5 s t u d e n t s f r o m e a c h c l a s s . \\ W e h a v e t h e f o l l o w i n g c a s e s . \\ (i) 5 s t u d e n t s f r o m X I c l a s s a n d 6 s t u d e n t s f r o m X I I c l a s s \\ (i i) 6 s t u d e n t s f r o m X I c l a s s a n d 5 s t u d e n t s f r o m X I I c l a s s \\ S o, n u m b e r o f w a y s o f s e l e c t i o n o f a t e a m o f 11 p l a y e r s \\ = {}_{20}{C_{5}} \times {}_{20}{C_{6}} + {}_{20}{C_{6}} \times {}_{20}{C_{5}} = 2 [{}_{20}{C_{5}} \times {}_{20}{C_{6}}] \\ H e n c e, t h e r e q u i r e d w a y s o f s e l e c t i o n = 2 [{}_{20}{C_{5}} \times {}_{20}{C_{6}}] \end{array}$

A group consists of 4 girls and 7 boys. In how many ways can a team of 5 members be selected if the team has:

(i) No girls.

(ii) At least one boy and one girl.

(iii) At least three girls.

This is a long answer type question as classified in NCERT Exemplar

$\begin{array}{l} W e h a v e 4 g i r l s a n d 7 b o y s a n d a t e a m o f 5 m e m b e r s i s t o b e s e l e c t e d . \\ (i) I f n o g i r l i s s e l e c t e d, t h e n a l l t h e 5 m e m b e r s a r e t o b e s e l e c t e d \\ o u t o f 7 b o y s i . e . {}_{7}{C_{5}} = \frac{7!}{5! 2!} = \frac{7 \times 6.5!}{5! \times 2} = 21 w a y s \\ (i i) W h e n a t l e a s t o n e b o y a n d o n e g i r l a r e t o b e s e l e c t e d, t h e n \\ N u m b e r o f w a y s = {}_{4}{C_{1}} \times {}_{7}{C_{4}} + {}_{4}{C_{2}} \times {}_{7}{C_{3}} + {}_{4}{C_{3}} \times {}_{7}{C_{2}} + {}_{4}{C_{4}} \times {}_{7}{C_{1}} \\ = 4 \times \frac{7 \times 6 \times 5 \times 4}{4 \times 3 \times 2 \times 1} + \frac{4 \times 3}{2 \times 1} \times \frac{7 \times 6 \times 5}{3 \times 2 \times 1} + 4 \times \frac{7 \times 6}{2 \times 1} + 1 \times 7 \\ = 4 \times 35 + 6 \times 35 + 4 \times 21 + 7 = 140 + 210 + 84 + 7 = 441 w a y s \\ (i i i) W h e n a t l e a s t 3 g i r l s a r e i n c l u d e d, t h e n \\ N u m b e r o f w a y s = {}_{4}{C_{3}} \times {}_{7}{C_{2}} + {}_{4}{C_{4}} \times {}_{7}{C_{1}} \\ = 4 \times \frac{7 \times 6}{2 \times 1} + 1 \times 7 = 84 + 7 = 91 w a y s \\ H e n c e, t h e r e q u i r e d n u m b e r o f w a y s a r e \\ (i) 21 w a y s (i i) 441 w a y s (i i i) 91 w a y s \end{array}$

Permutations and Combinations Objective Type Questions

1. If ⁿC₁₂ = ⁿC₈, then n is equal to:

(a) 20

(b) 12

(c) 6

(d) 30

Sol.

\begin{array}{l} G i v e n t h a t : {}_{n}{C_{12}} = {}_{n}{C_{8}} \\ \Rightarrow {}_{n}{C_{n - 12}} = {}_{n}{C_{8}} [∵ {}_{n}{C_{r}} = {}_{n}{C_{n - r}}] \\ \Rightarrow n - 12 = 8 \\ \Rightarrow n = 8 + 12 = 20 \\ H e n c e, t h e c o r r e c t o p t i o n i s (a) . \end{array}

2.The number of possible outcomes when a coin is tossed 6 times is:

(a) 36

(b) 64

(c) 12

(d) 32

Sol.

\begin{array}{l} W e k n o w t h a t a c o i n h a s H e a d a n d T a i l (H, T) \\ ∴ W h e n a c o i n i s t o s s e d 6 t i m e s, t h e n t h e \\ P o s s i b l e o u t c o m e = 2^{6} = 64 \\ H e n c e, t h e c o r r e c t o p t i o n i s (b) . \end{array}

Commonly asked questions

The total number of 9-digit numbers which have all different digits is:

(a) 10!

(b) 9!

(c) 9 × 9!

(d) 10 × 10!

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l} W e h a v e t o f o r m 9 d i g i t n u m b e r s f r o m 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 a n d w e k n o w t h a t 0 c a n n o t b e p u t \\ o n e x t r e m e l y l e f t p l a c e . \\ S o, f i r s t p l a c e f r o m t h e l e f t s i d e c a n b e f i l l e d i n 9 w a y s . \\ N o w r e p e t i t i o n i s n o t a l l o w e d . S o, t h e r e m a i n i n g 8 p l a c e s c a n b e f i l l e d i n 9! \\ ∴ T h e r e q u i r e d n u m b e r o f w a y s = 9 \times 9! \\ H e n c e, t h e c o r r e c t o p t i o n i s (c) . \end{array}$

The number of words that can be formed out of the letters of the word "ARTICLE" so that vowels occupy the even places is:

(a) 1440

(b) 144

(c) 7!

(d) ⁴C₄ × ³C₃

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l} T o t a l n u m b e r o f l e t t e r s i n t h e' A R T I C L E' i s 7 o u t w h i c h A, E, I a r e v o w e l s a n d \\ R, T, C, L a r e c o n s o n a n t s \\ G i v e n t h a t v o w e l s o c c u p y e v e n p l a c e \\ ∴ P o s s i b l e a r r a n g e m e n t c a n b e s h o w n a s b e l o w \\ C, V, C, V, C, V, C i . e . o n 2 n d, 4 t h a n d 6 t h p l a c e s \\ T h e r e f o r e, n u m b e r o f a r r a n g e m e n t = {}_{3}{P_{3}} = 3! = 6 w a y s \\ N o w c o n s o n a n t s c a n b e p l a c e d a t 1, 3, 5 a n d 7 t h p l a c e \\ ∴ N u m b e r o f a r r a n g e m e n t = {}_{4}{P_{4}} = 4! = 24 \\ S o, t h e t o t a l n u m b e r o f a r r a n g e m e n t s = 6 \times 24 = 144 \\ H e n c e, t h e c o r r e c t o p t i o n i s (b) . \end{array}$

Given 5 different green dyes, four different blue dyes, and three different red dyes, the number of combinations of dyes that can be chosen, taking at least one green and one blue dye, is:

(a) 3600

(b) 3720

(c) 3800

(d) 3600

[Hint: Possible numbers of choosing or not choosing 5 green dyes, 4 blue dyes, and 3 red dyes are $2^{5}, 2^{4}, 2^{3}$ , respectively.]

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l} P o s s i b l e n u m b e r o f choosing 5 d i f f e r e n t g r e e n d y e s = 2^{5} \\ P o s s i b l e n u m b e r o f choosing 4 b l u e d y e s = 2^{4} \\ a n d p o s s i b l e n u m b e r o f choosing 3 r e d d y e s = 2^{3} \\ I f a t l e a s t o n e b l u e a n d o n e g r e e n d y e s a r e s e l e c t e d t h e n t h e t o t a l n u m b e r o f s e l e c t i o n \\ = (2^{5} - 1) \times (2^{4} - 1) \times 2^{3} = 31 \times 15 \times 8 = 3720 \\ H e n c e, t h e c o r r e c t o p t i o n i s (b) . \end{array}$

If ⁿC₁₂ = ⁿC₈, then n is equal to:

(a) 20

(b) 12

(c) 6

(d) 30

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l} G i v e n t h a t : {}_{n}{C_{12}} = {}_{n}{C_{8}} \\ \Rightarrow {}_{n}{C_{n - 12}} = {}_{n}{C_{8}} [? {}_{n}{C_{r}} = {}_{n}{C_{n - r}}] \\ \Rightarrow n - 12 = 8 \\ \Rightarrow n = 8 + 12 = 20 \\ H e n c e, t h e c o r r e c t o p t i o n i s (a) . \end{array}$

The number of possible outcomes when a coin is tossed 6 times is:

(a) 36

(b) 64

(c) 12

(d) 32

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l} W e k n o w t h a t a c o i n h a s H e a d a n d T a i l (H, T) \\ ∴ W h e n a c o i n i s t o s s e d 6 t i m e s, t h e n t h e \\ P o s s i b l e o u t c o m e = 2^{6} = 64 \\ H e n c e, t h e c o r r e c t o p t i o n i s (b) . \end{array}$

The number of different four-digit numbers that can be formed with the digits 2, 3, 4, 7 and using each digit only once is:

(a) 120

(b) 96

(c) 24

(d) 100

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l} F o u r - d i g i t n u m b e r s a r e t o b e f o r m e d f r o m t h e d i g i t s 2, 3, 4, 7 w i t h o u t r e p e t i t i o n \\ S o, t h e r e q u i r e d 4 - d i g i t n u m b e r s = {}_{4}{P_{4}} = 4! = 4 \times 3 \times 2 \times 1 = 24 \\ H e n c e, t h e c o r r e c t o p t i o n i s (c) . \end{array}$

The sum of the digits in the unit place of all the numbers formed with the help of 3, 4, 5, and 6 taken all at a time is:

(a) 432

(b) 108

(c) 36

(d) 18

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l} I f w e f i x 3 a t u n i t p l a c e, t h e n t h e t o t a l p o s s i b l e o u t c o m e s = 3! \\ I f w e f i x 4, 5 a n d 6 a t u n i t p l a c e, t h i s i s e a c h c a s e, t o t a l p o s s i b l e n u m b e r s a r e 3! \\ Required s u m o f u n i t d i g i t s o f a l l s u c h n u m b e r s i s \\ = 3 \times 3! + 4 \times 3! + 5 \times 3! + 6 \times 3! = (3 + 4 + 5 + 6) \times 3! \\ = 18 \times 3! = 18 \times 3 \times 2 \times 1 = 108 \\ H e n c e, t h e c o r r e c t o p t i o n i s (b) . \end{array}$

The total number of words formed by 2 vowels and 3 consonants taken from 4 vowels and 5 consonants is equal to:

(a) 60

(b) 120

(c) 7200

(d) 720

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l} G i v e n t h a t t o t a l n u m b e r s o f v o w e l s = 4 \\ a n d t o t a l n u m b e r s o f c o n s o n a n t s = 5 \\ T o t a l n u m b e r o f w o r d s f o r m e d b y 2 v o w e l s a n d 3 c o n s o n a n t s \\ = {}_{4}{C_{2}} \times {}_{5}{C_{3}} = \frac{4!}{2! 2!} \times \frac{5!}{3! 2!} = \frac{4 \times 3 \times 2!}{2 \times 1 \times 2!} \times \frac{5 \times 4 \times 3!}{3! \times 2} \\ = 6 \times 10 = 60 \\ N o w p e r m u t a t i o n o f 2 v o w e l s a n d 3 c o n s o n a n t s = 5! \\ = 5 \times 4 \times 3 \times 2 \times 1 = 120 \\ S o, t h e t o t a l n u m b e r o f w o r d s = 60 \times 120 = 7200 . \\ H e n c e, t h e c o r r e c t o p t i o n i s (c) . \end{array}$

A five-digit number divisible by 3 is to be formed using the numbers 0, 1, 2, 3, 4, and 5 without repetitions. The total number of ways this can be done is:

(a) 216

(b) 600

(c) 240

(d) 3125

[Hint: 5-digit numbers can be formed using digits 0, 1, 2, 4, 5 or by using digits 1, 2, 3, 4, 5 since the sum of digits in these cases is divisible by 3.]

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l} W e k n o w t h a t a n u m b e r i s d i v i s i b l e b y 3 w h e n t h e s u m o f i t s d i g i t s i s d i v i s i b l e b y 3 . \\ I f w e t a k e t h e d i g i t s 0, 1, 2, 4, 5, t h e n t h e s u m o f t h e d i g i t s = 0 + 1 + 2 + 4 + 5 = 12 w h i c h i s \\ d i v i s i b l e b y 3 . \\ S o, t h e 5 d i g i t n u m b e r s using t h e d i g i t s 0, 1, 2, 4 a n d 5 \\ \underset{4}{T T H} \underset{4}{T H} \underset{3}{H} \underset{2}{T} \underset{1}{O} \\ = 4 \times 4 \times 3 \times 2 \times 1 = 96 \\ a n d i f w e t a k e t h e d i g i t s 1, 2, 3, 4, 5, t h e n t h e i r s u m \\ = 1 + 2 + 3 + 4 + 5 = 15 w h i c h i s d i v i s i b l e b y 3 . \\ S o, t h e 5 d i g i t n u m b e r s c a n b e f o r m e d using t h e d i g i t s 1, 2, 3, 4, 5 i s 5! w a y s \\ = 5 \times 4 \times 3 \times 2 \times 1 = 120 \\ T o t a l n u m b e r o f w a y s = 96 + 120 = 216 \\ H e n c e, t h e c o r r e c t o p t i o n i s (a) . \end{array}$

Everybody in a room shakes hands with everybody else. The total number of handshakes is 66. The total number of persons in the room is:

(a) 11

(b) 12

(c) 13

(d) 14

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l} L e t t h e t o t a l n u m b e r o f p e r s o n s i n a r o o m b e n since, t w o p e r s o n s m a k e 1 h a n d s h a k e . \\ ∴ T h e n u m b e r o f h a n d s h a k e s = {}_{n}{C_{2}} \\ S o, {}_{n}{C_{2}} = 66 \\ \Rightarrow \frac{n!}{2! (n - 2)!} = 66 \Rightarrow \frac{n (n - 1) (n - 2)!}{2 \times 1 \times (n - 2)!} = 66 \\ \Rightarrow \frac{n (n - 1)}{2} = 66 \Rightarrow n^{2} - n = 132 \\ \Rightarrow n^{2} - n - 132 = 0 \Rightarrow n^{2} - 12 n + 11 n - 132 = 0 \\ \Rightarrow n (n - 12) + 11 (n - 12) = 0 \Rightarrow (n - 12) (n + 11) = 0 \\ \Rightarrow n - 12 = 0, n + 11 = 0 \Rightarrow n = 12, n = - 11 \\ \Rightarrow n = 12 (? n \neq - 11) \\ H e n c e, t h e c o r r e c t o p t i o n i s (b) . \end{array}$

The number of triangles that are formed by choosing the vertices from a set of 12 points, seven of which lie on the same line, is:

(a) 105

(b) 15

(c) 75

(d) 185

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l} T h e t o t a l n u m b e r o f t r i a n g l e s f o r m e d f r o m 12 points t a k i n g 3 a t a t i m e = {}_{12}{C_{3}} . \\ B u t g i v e n t h a t o u t o f 12 points, 7 a r e c o l l i n e a r \\ S o, t h e s e s e v e n points w i l l f o r m n o t r i a n g l e . \\ ∴ T h e r e q u i r e d n u m b e r o f t r i a n g l e s = {}_{12}{C_{3}} - {}_{7}{C_{3}} \\ \Rightarrow \frac{12!}{3! 9!} - \frac{7!}{3! 4!} = \frac{12 \times 11 \times 10 \times 9!}{3 \times 2 \times 1 \times 9!} - \frac{7 \times 6 \times 5 \times 4!}{3 \times 2 \times 1 \times 4!} \\ \Rightarrow \frac{12 \times 11 \times 10}{3 \times 2} - \frac{7 \times 6 \times 5}{3 \times 2} = 220 - 35 = 185 \\ H e n c e, t h e c o r r e c t o p t i o n i s (d) . \end{array}$

The number of parallelograms that can be formed from a set of four parallel lines intersecting another set of three parallel lines is:

(a) 6

(b) 18

(c) 12

(d) 9

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l} W e k n o w t h a t t o f o r m a parallelogram, w e r e q u i r e a p a i r o f l i n e s f r o m a s e t o f 4 l i n e s a n d \\ a n o t h e r p a i r o f l i n e s f r o m a n o t h e r s e t o f 3 l i n e s \\ ∴ T h e r e q u i r e d n u m b e r o f parallelograms = {}_{4}{C_{2}} \times {}_{3}{C_{2}} = 6 \times 3 = 18 \\ H e n c e, t h e c o r r e c t o p t i o n i s (b) . \end{array}$

The number of ways in which a team of eleven players can be selected from 22 players always including 2 of them and excluding 4 of them is:

(a) ¹⁶C₁₁

(b) ¹⁶C₅

(c) ¹⁶C₉

(d) ²⁰C₉

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l} T o t a l n u m b e r o f p l a y e r s = 22 \\ 2 p l a y e r s a r e a l w a y s i n c l u d e d a n d 4 a r e a l w a y s e x c l u d i n g o r n e v e r i n c l u d e d = 22 - 2 - 4 = 16 \\ ∴ Required n u m b e r o f s e l e c t i o n = {}_{16}{C_{9}} \\ H e n c e, t h e c o r r e c t o p t i o n i s (c) . \end{array}$

The number of 5-digit telephone numbers having at least one of their digits repeated is:

(a) 90,000

(b) 10,000

(c) 30,240

(d) 69,760

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l} T o t a l n u m b e r o f 5 - d i g i t t e l e p h o n e n u m b e r i f a l l t h e d i g i t s a r e r e p e a t e d = {(10)}^{5} \\ [? d i g i t s a r e f r o m 0 t o 9] \\ I f d i g i t s a r e n o t r e p e a t e d, t h e n 5 - d i g i t t e l e p h o n e s, c a n b e f o r m e d i n {}_{10}{P_{5}} w a y s \\ ∴ Required n u m b e r o f w a y s = {(10)}^{5} - {}_{10}{P_{5}} \\ = 100000 - \frac{10!}{(10 - 5)!} = 100000 - \frac{10 \times 9 \times 8 \times 7 \times 6 \times 5!}{5!} \\ = 100000 - 30240 = 69760 \\ H e n c e, t h e c o r r e c t o p t i o n i s (d) . \end{array}$

The number of ways in which we can choose a committee from four men and six women so that the committee includes at least two men and exactly twice as many women as men is:

(a) 94

(b) 126

(c) 128

(d) None

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l} N u m b e r o f m e n = 4 \\ N u m b e r o f w o m e n = 6 \\ W e a r e g i v e n t h a t t h e c o m m i t t e e i n c l u d e s 2 m e n a n d e x a c t l y t w i c e a s m a n y w o m e n a s m e n . \\ T h u s, t h e p o s s i b l e s e l e c t i o n c a n b e \\ 2 m e n a n d 4 w o m e n a n d 3 m e n a n d 6 w o m e n . \\ S o, t h e n u m b e r o f c o m m i t t e e = {}_{4}{C_{2}} \times {}_{6}{C_{4}} + {}_{4}{C_{3}} \times {}_{6}{C_{6}} \\ = 6 \times 15 + 4 \times 1 = 90 + 4 = 94 \\ H e n c e, t h e c o r r e c t o p t i o n i s (a) . \end{array}$

Permutations and Combinations Fill in the Blanks Type Questions

1. If ⁿp_r = 840, ⁿC_r = 35, then $r =$ ______.

Sol.

\begin{array}{l} G i v e n t h a t {}_{n}{P_{r}} = 840 a n d {}_{n}{C_{r}} = 35 \\ \Rightarrow \frac{n!}{(n - r)!} = 840 \dots (i) a n d \frac{n!}{r! (n - r)!} = 35 \dots (i i) \\ D i v i d i n g e q n . (i) b y (i i) w e g e t \\ \frac{\frac{n!}{(n - r)!}}{\frac{n!}{r! (n - r)!}} = \frac{840}{35} \Rightarrow \frac{n!}{(n - r)!} \times \frac{r! (n - r)!}{n!} = 24 \\ \Rightarrow r! = 24 \Rightarrow r! = 4 \times 3 \times 2 \times 1 \\ \Rightarrow r! = 4! ∴ r = 4 \\ H e n c e, t h e v a l u e o f t h e f i l l e r i s 4 . \end{array}

2.¹⁵C₈ + ¹⁵C₉ – ¹⁵C₆ – ¹⁵C₇ = ______.

Sol.

\begin{array}{l} {}_{15}{C_{8}} + {}_{15}{C_{9}} - {}_{15}{C_{6}} - {}_{15}{C_{7}} = {}_{15}{C_{15 - 8}} + {}_{15}{C_{15 - 9}} - {}_{15}{C_{6}} - {}_{15}{C_{7}} [∵ {}_{n}{C_{r}} = {}_{n}{C_{n - r}}] \\ = {}_{15}{C_{7}} + {}_{15}{C_{6}} - {}_{15}{C_{6}} - {}_{15}{C_{7}} = 0 \\ H e n c e, t h e v a l u e o f t h e f i l l e r i s 0 . \end{array}

Commonly asked questions

If ⁿp_r = 840, ⁿC_r = 35, then $r =$ ______.

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l} G i v e n t h a t {}_{n}{P_{r}} = 840 a n d {}_{n}{C_{r}} = 35 \\ \Rightarrow \frac{n!}{(n - r)!} = 840 \dots (i) a n d \frac{n!}{r! (n - r)!} = 35 \dots (i i) \\ D i v i d i n g e q n . (i) b y (i i) w e g e t \\ \frac{\frac{n!}{(n - r)!}}{\frac{n!}{r! (n - r)!}} = \frac{840}{35} \Rightarrow \frac{n!}{(n - r)!} \times \frac{r! (n - r)!}{n!} = 24 \\ \Rightarrow r! = 24 \Rightarrow r! = 4 \times 3 \times 2 \times 1 \\ \Rightarrow r! = 4! ∴ r = 4 \\ H e n c e, t h e v a l u e o f t h e f i l l e r i s 4 . \end{array}$

¹⁵C₈ + ¹⁵C₉ – ¹⁵C₆ – ¹⁵C₇ = ______.

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l} {}_{15}{C_{8}} + {}_{15}{C_{9}} - {}_{15}{C_{6}} - {}_{15}{C_{7}} = {}_{15}{C_{15 - 8}} + {}_{15}{C_{15 - 9}} - {}_{15}{C_{6}} - {}_{15}{C_{7}} [? {}_{n}{C_{r}} = {}_{n}{C_{n - r}}] \\ = {}_{15}{C_{7}} + {}_{15}{C_{6}} - {}_{15}{C_{6}} - {}_{15}{C_{7}} = 0 \\ H e n c e, t h e v a l u e o f t h e f i l l e r i s 0 . \end{array}$

The number of permutations of $n$ different objects, taken $r$ at a time, when repetitions are allowed, is ______.

This is a Fill in the blanks Type Questions as classified in NCERT Exemplar

$N u m b e r o f p e r m u t a t i o n o f n d i f f e r e n t o b j e c t s, t a k e n r a t a t i m e i s n^{r} .$

The number of different words that can be formed from the letters of the word INTERMEDIATE such that two vowels never come together is ______.
[Hint: Number of ways of arranging 6 consonants of which two are alike is $\frac{6!}{2!}$ and number of ways of arranging vowels $7 P 6 = \frac{3! \times 2!}{1! \times 2!}$ .]

This is a Fill in the blanks Type Questions as classified in NCERT Exemplar

$\begin{array}{l} T o t a l n u m b e r o f w o r d s i s I N T E R M E D I A T E = 12 \\ w h i c h h a v e 6 v o w e l s a n d 6 c o n s o n a n t s \\ I f t w o v o w e l s n e v e r c o m e t o g e t h e r t h e n w e c a n a r r a n g e a s u n d e r \\ V C V C V C V C V C V C V \\ H e r e, v o w e l s a r e I E E I A E w h e r e 2 I' s a n d 3 E' s a r e t h e r e . \\ ∴ N u m b e r o f w a y s o f a r r a n g i n g v o w e l s = \frac{7!}{3! 2!} = 420 \\ c o n s o n a n t s a r e N T R M D T w h e r e 2 T' s a r e t h e r e . \\ ∴ N u m b e r o f w a y s o f a r r a n g i n g c o n s o n a n t s = \frac{6!}{2!} = \frac{6.5.4.3.2!}{2!} = 360 \\ S o, t h e t o t a l n u m b e r o f w o r d s a r e = 420 \times 360 = 151200 . \\ H e n c e, t h e v a l u e o f t h e f i l l e r i s 151200 . \end{array}$

Three balls are drawn from a bag containing 5 red, 4 white, and 3 black balls. The number of ways in which this can be done if at least 2 are red is ______.

This is a Fill in the blanks Type Questions as classified in NCERT Exemplar

$\begin{array}{l} W e h a v e 5 r e d, 4 w h i t e a n d 3 b l a c k b a l l s o u t o f w h i c h a t l e a s t 2 r e d b a l l s a r e t o b e d r a w n . \\ ∴ N u m b e r o f w a y s = {}_{5}{C_{2}} \times {}_{7}{C_{1}} + {}_{5}{C_{3}} \\ = 10 \times 7 + 10 = 70 + 10 = 80 \\ H e n c e, t h e v a l u e o f t h e f i l l e r i s 80 . \end{array}$

The number of six-digit numbers, all digits of which are odd, is ______.

This is a Fill in the blanks Type Questions as classified in NCERT Exemplar

$\begin{array}{l} O u t o f t h e d i g i t s 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 t h e o d d d i g i t s a r e 1, 3, 5, 7, 9 . \\ T h e r f o r e, n u m b e r o f 6 d i g i t s n u m b e r s = {(5)}^{6} \\ H e n c e, t h e v a l u e o f t h e f i l l e r i s {(5)}^{6} . \end{array}$

In a football championship, 153 matches were played. Every two teams played one match with each other. The number of teams participating in the championship is ______.

This is a Fill in the blanks Type Questions as classified in NCERT Exemplar

$\begin{array}{l} L e t t h e n u m b e r o f p a r t i c i p a t i n g t e a m s b e n . \\ G i v e n t h a t e v e r y t w o t e a m s p l a y e d o n e m a t c h w i t h e a c h o t h e r . \\ ∴ T o t a l n u m b e r o f m a t c h e s p l a y e d = {}_{n}{C_{2}} \\ S o {}_{n}{C_{2}} = 153 \\ \Rightarrow \frac{n (n - 1)}{2} = 153 \Rightarrow n^{2} - n = 306 \\ \Rightarrow n^{2} - n - 306 = 0 \Rightarrow n^{2} - 18 n + 17 n - 306 = 0 \\ \Rightarrow n (n - 18) + 17 (n - 18) = 0 \Rightarrow (n - 18) (n + 17) = 0 \\ \Rightarrow n - 18 = 0 a n d n + 17 = 0 \Rightarrow n = 18, n \neq - 17 \\ H e n c e, t h e v a l u e o f t h e f i l l e r i s 18 . \end{array}$

The total number of ways in which six ‘+’ and four ‘–’ signs can be arranged in a line such that no two signs ‘–’ occur together is ______.

This is a Fill in the blanks Type Questions as classified in NCERT Exemplar

$\begin{array}{l} T h e f o l l o w i n g m a y b e t h e a r r a n g e m e n t o f (-) a n d (+) \\ (-) (+) (-) (+) (-) (+) (-) (+) (-) (+) (-) (+) (-) \\ T h e r e f o r e,' +' s i g n c a n b e a r r a n g e d o n l y i s 1 w a y b e c a u s e a l l a r e i d e n t i c a l . \\ a n d 4 (-) s i g n s c a n b e a r r a n g e d a t 7 p l a c e s i n {}_{7}{C_{4}} w a y s \\ ∴ T o t a l n u m b e r o f w a y s = {}_{7}{C_{4}} \times 1 = \frac{7 \times 6 \times 5 \times 4}{4 \times 3 \times 2 \times 1} \times 1 = 35 w a y s \\ H e n c e, t h e v a l u e o f t h e f i l l e r i s 35 . \end{array}$

A committee of 6 is to be chosen from 10 men and 7 women so as to contain at least 3 men and 2 women. In how many different ways can this be done if two particular women refuse to serve on the same committee?
[Hint: At least 3 men and 2 women: The number of ways = ¹⁰C₃ × ¹⁰C₃ + ¹⁰C₄ × ⁷C₂. For 2 particular women to be always there: the number of ways = ¹⁰C₄ × ¹⁰C₃ × ⁵C₁. The total number of committees when two particular women are never together = Total – together.]

This is a Fill in the blanks Type Questions as classified in NCERT Exemplar

$\begin{array}{l} W e h a v e 10 m e n a n d 7 w o m e n o u t o f w h i c h a c o m m i t t e e o f 6 i s t o b e f o r m e d w h i c h c o n t a i n \\ a t l e a s t 3 m e n a n d 2 w o m e n \\ T h e r e f o r e, N u m b e r o f w a y s = {}_{10}{C_{3}} \times {}_{7}{C_{3}} + {}_{10}{C_{4}} \times {}_{7}{C_{2}} \\ = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} \times \frac{7 \times 6 \times 5}{3 \times 2 \times 1} + \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} \times \frac{7 \times 6}{2 \times 1} \\ = 120 \times 35 + 210 \times 21 = 4200 + 4410 = 8610 \\ I f 2 p a r t i c u l a r w o m e n t o b e a l w a y s p r e s e n t, t h e n t h e n u m b e r o f w a y s = {}_{10}{C_{4}} \times {}_{5}{C_{0}} + {}_{10}{C_{3}} \times {}_{5}{C_{1}} \\ = \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} \times 1 + \frac{10 \times 9 \times 8}{3 \times 2 \times 1} \times 5 = 210 + 120 \times 5 \\ = 210 + 600 = 810 \\ ∴ T o t a l n u m b e r o f c o m m i t t e e = 8610 - 810 = 7800 \\ H e n c e, t h e v a l u e o f t h e f i l l e r i s 7800 . \end{array}$

A box contains 2 white balls, 3 black balls, and 4 red balls. The number of ways three balls can be drawn from the box if at least one black ball is to be included in the draw is ______.

This is a Fill in the blanks Type Questions as classified in NCERT Exemplar

$\begin{array}{l} W e h a v e 2 w h i t e, 3 b l a c k a n d 4 r e d b a l l s \\ I t i s g i v e n t h a t a t l e a s t 1 b l a c k b a l l i s t o b e i n c l u d e d . \\ ∴ Required n u m b e r o f w a y s = {}_{3}{C_{1}} \times {}_{6}{C_{2}} + {}_{3}{C_{2}} \times {}_{6}{C_{1}} + {}_{3}{C_{3}} \\ = 3 \times 15 + 3 \times 6 + 1 = 45 + 18 + 1 = 64 \\ H e n c e, t h e v a l u e o f t h e f i l l e r i s 64 . \end{array}$

Permutations and Combinations True or False Type Questions

1. There are 12 points in a plane of which 5 points are collinear, then the number of lines obtained by joining these points in pairs is ¹²C₂ - ⁵C₂.

Sol.

\begin{array}{l} Required n u m b e r o f l i n e s = {}_{12}{C_{2}} - {}_{5}{C_{2}} + 1 \\ H e n c e, t h e g i v e n s t a t e m e n t i s' F a l s e' \end{array}

2. Three letters can be posted in five letterboxes in 35 ways.

Sol.

\begin{array}{l} G i v e n t h a t 3 l e t t e r s a r e t o b e p o s t e d i n 5 l e t t e r b o x e s \\ ∴ Required n u m b e r o f w a y s = 5^{3} = 125 \\ H e n c e, t h e g i v e n s t a t e m e n t i s' F a l s e' \end{array}

Commonly asked questions

Eighteen guests are to be seated, half on each side of a long table. Four particular guests desire to sit on one particular side and three others on the other side of the table. The number of ways in which the seating arrangements can be made is $\frac{11! (9!) (9!)}{5! 6!}$ .

This is a True or False Type Questions as classified in NCERT Exemplar

$\begin{array}{l} W h e n 4 g u e s t s s i t c a n o n e s i d e a n d 3 o n t h e o t h e r s i d e, w e h a v e t o s e l e c t o u t 11.5 s i t o n e s i d e \\ a n d 6 s i t o n t h e o t h e r s i d e . \\ N o w, r e m a i n i n g s e l e c t i n g o n o n e h a l f s i d e = {}_{18 - 4 - 3}{C_{5}} = {}_{11}{C_{5}} \\ a n d t h e o t h e r h a l f s i d e = {}_{11 - 5}{C_{6}} = {}_{6}{C_{6}} \\ S o, t h e t o t a l a r r a n g e m e n t s = {}_{11}{C_{5}} \times 9! \times {}_{6}{C_{6}} \times 9! \\ = \frac{11!}{5! 6!} \times 9! \times 1 \times 9! = \frac{11!}{5! 6!} (9!) (9!) \\ H e n c e, t h e g i v e n s t a t e m e n t i s' T r u e' . \end{array}$

There are 12 points in a plane of which 5 points are collinear, then the number of lines obtained by joining these points in pairs is ¹²C₂ - ⁵C₂.

This is a True or False Type Questions as classified in NCERT Exemplar

$\begin{array}{l} Required n u m b e r o f l i n e s = {}_{12}{C_{2}} - {}_{5}{C_{2}} + 1 \\ H e n c e, t h e g i v e n s t a t e m e n t i s' F a l s e' \end{array}$

In the permutations of $n$ things, $r$ taken together, the number of permutations in which $m$ particular things occur together is $n - m P_{r} - m \times r P m$ .

This is a True or False Type Questions as classified in NCERT Exemplar

$\begin{array}{l} A r r a n g e m e n t o f n t h i n g s, r t a k e n a t a t i m e i n w h i c h m t h i n g s o c c u r t o g e t h e r . \\ S o, n u m b e r o f o b j e c t e x c l u d i n g m o b j e c t = (r - m) \\ H e r e, w e f i r s t a r r a n g e (r - m + 1) o b j e c t \\ ∴ N u m b e r o f a r r a n g e m e n t s = (r - m + 1)! \\ m o b j e c t s c a n b e a r r a n g e d i n m! w a y s \\ S o, t h e r e q u i r e d n u m b e r o f a r r a n g e m e n t s = (r - m + 1)! \times m! \\ H e n c e, t h e g i v e n s t a t e m e n t i s' F a l s e' . \end{array}$

In a steamer, there are stalls for 12 animals, and there are horses, cows, and calves (not less than 12 each) ready to be shipped. They can be loaded in 312 ways.

This is a True or False Type Questions as classified in NCERT Exemplar

$\begin{array}{l} T h e r e a r e 3 t y p e s o f a n i m a l s h o r s e s, c o w s a n d c a l v e s n o t l e s s t h a n 12 e a c h . \\ S o, n u m b e r o f w a y s o f l o a d i n g = 3^{12} \\ H e n c e, t h e g i v e n s t a t e m e n t i s' T r u e' . \end{array}$

Three letters can be posted in five letterboxes in 35 ways.

This is a True or False Type Questions as classified in NCERT Exemplar

$\begin{array}{l} G i v e n t h a t 3 l e t t e r s a r e t o b e p o s t e d i n 5 l e t t e r b o x e s \\ ∴ Required n u m b e r o f w a y s = 5^{3} = 125 \\ H e n c e, t h e g i v e n s t a t e m e n t i s' F a l s e' \end{array}$

If some or all of $n$ objects are taken at a time, the number of combinations is $2^{n} - 1$ .

This is a True or False Type Questions as classified in NCERT Exemplar

$\begin{array}{l} W h e n s o m e o r a l l o b j e c t s, t a k e n a t a t i m e, t h e n t h e n u m b e r o f s e l e c t i o n w i l l b e \\ {}_{n}{C_{1}} + {}_{n}{C_{2}} + {}_{n}{C_{3}} + \dots + {}_{n}{C_{n}} = 2^{n} - 1 [? {}_{n}{C_{0}} + {}_{n}{C_{1}} + {}_{n}{C_{2}} + {}_{n}{C_{3}} + \dots + {}_{n}{C_{n}} = 2^{n}] \\ H e n c e, t h e g i v e n s t a t e m e n t i s' T r u e' . \end{array}$

There will be only 24 selections containing at least one red ball out of a bag containing 4 red and 5 black balls. It is given that the balls of the same colour are identical.

This is a True or False Type Questions as classified in NCERT Exemplar

$\begin{array}{l} W e h a v e 4 r e d a n d 5 b l a c k b a l l s i n a b o x a n d a t l e a s t o n e r e d b a l l i s t o b e d r a w n \\ ∴ N u m b e r o f s e l e c t i o n = [(4 + 1) (5 + 1) - 1] - 5 = [5 \times 6 - 1] - 5 \\ 29 - 5 = 24 \\ H e n c e, t h e g i v e n s t a t e m e n t i s' T r u e' . \end{array}$

A candidate is required to answer 7 questions out of 12 questions which are divided into two groups, each containing 6 questions. He is not permitted to attempt more than 5 questions from either group. He can choose the seven questions in 650 ways.

This is a True or False Type Questions as classified in NCERT Exemplar

To fill 12 vacancies there are 25 candidates, of which 5 are from scheduled castes. If 3 of the vacancies are reserved for scheduled caste candidates while the rest are open to all, the number of ways in which the selection can be made is ⁵C₃ × ²⁰.

This is a True or False Type Questions as classified in NCERT Exemplar

$\begin{array}{l} N u m b e r o f w a y s t o s e l e c t 3 s c h e d u l e d c a s t e c a n d i d a t e o u t o f 5 = {}_{5}{C_{3}} \\ W e h a v e t o s e l e c t 9 o t h e r c a n d i d a t e o u t o f 22 . \\ S o t h e n u m b e r o f w a y s = {}_{22}{C_{9}} \\ Required n u m b e r o f s e l e c t i o n = {}_{5}{C_{3}} \times {}_{22}{C_{9}} \\ H e n c e, t h e g i v e n s t a t e m e n t i s' F a l s e' . \end{array}$

Permutations and Combinations Matching Type Questions

1. There are 3 books on Mathematics, 4 on Physics, and 5 on English. How many different collections can be made such that each collection consists of:

C₁ | C₂

(a) One book of each subject; | (i) 3968

(b) At least one book of each subject: | (ii) 60

(c) At least one book of English: | (iii) 3255

Sol.

\begin{array}{l} W e h a v e 3 b o o k s o f M a t h e m a t i c s, 4 o f P h y s i c s a n d 5 o n E n g l i s h \\ (a) O n e b o o k o f e a c h s u b j e c t = {}_{3}{C_{1}} \times {}_{4}{C_{1}} \times {}_{5}{C_{1}} = 3 \times 4 \times 5 = 60 \\ (b) A t l e a s t o n e b o o k o f e a c h s u b j e c t = (2^{3} - 1) \times (2^{4} - 1) \times (2^{5} - 1) = 7 \times 15 \times 31 = 3255 \\ (c) A t l e a s t o n e b o o k o f E n g l i s h = (2^{5} - 1) \times 2^{7} = 31 \times 128 = 3968 \\ H e n c e, t h e r e q u i r e d m a t c h i n g i s \\ (a) \leftrightarrow (i i), (b) \leftrightarrow (i i i) a n d (c) \leftrightarrow (i) \end{array}

2.Five boys and five girls form a line. Find the number of ways of making the seating arrangement under the following condition

C₁ | C₂

(a) Boys and girls alternate: | (i) 5! × 6!

(b) No two girls sit together: | (ii) 10! – 5! 6!

(c) All the girls sit together: | (iii) (5!)² + (5!)²

(d) All the girls are never together: | (iv) 2! 5! 5!

Sol.

\begin{array}{l} (a) T o t a l n u m b e r o f a r r a n g e m e n t w h e n b o y s a n d g i r l s a l t e r n a t e : = {(5!)}^{2} + {(5!)}^{2} \\ (b) N o t w o g i r l s s i t t o g e t h e r : = 5! 6! \\ (c) A l l t h e g i r l s s i t t o g e t h e r : = 2! 5! 5! \\ (d) A l l t h e g i r l s n e v e r s i t t o g e t h e r : = 10! - 5! 6! \\ H e n c e, t h e r e q u i r e d m a t c h i n g i s \\ (a) \leftrightarrow (i i i), (b) \leftrightarrow (i), (c) \leftrightarrow (i v) a n d (d) \leftrightarrow (i i) \end{array}

Commonly asked questions

Five boys and five girls form a line. Find the number of ways of making the seating arrangement under the following condition

C₁ | C₂

(a) Boys and girls alternate: | (i) 5! × 6!

(b) No two girls sit together: | (ii) 10! – 5! 6!

(c) All the girls sit together: | (iii) (5!)² + (5!)²

(d) All the girls are never together: | (iv) 2! 5! 5!

This is a matching answer type question as classified in NCERT Exemplar

$\begin{array}{l} (a) T o t a l n u m b e r o f a r r a n g e m e n t w h e n b o y s a n d g i r l s a l t e r n a t e : = {(5!)}^{2} + {(5!)}^{2} \\ (b) N o t w o g i r l s s i t t o g e t h e r : = 5! 6! \\ (c) A l l t h e g i r l s s i t t o g e t h e r : = 2! 5! 5! \\ (d) A l l t h e g i r l s n e v e r s i t t o g e t h e r : = 10! - 5! 6! \\ H e n c e, t h e r e q u i r e d m a t c h i n g i s \\ (a) \leftrightarrow (i i i), (b) \leftrightarrow (i), (c) \leftrightarrow (i v) a n d (d) \leftrightarrow (i i) \end{array}$

There are 10 professors and 20 lecturers, out of whom a committee of 2 professors and 3 lecturers is to be formed. Find:

C₁ | C₂

(a) In how many ways can the committee be formed: | (i) ¹⁰C₂ × ¹⁹C₃

(b) In how many ways a particular professor is included: | (ii) ¹⁰C₂ × ¹⁹C₂

(c) In how many ways a particular lecturer is included: | (iii) ⁹C₁ × ²⁰C₃

(d) In how many ways a particular lecturer is excluded: | (iv) ¹⁰C₂ × ²⁰C₃

This is a matching answer type question as classified in NCERT Exemplar

$\begin{array}{l} (a) W e h a v e t o s e l e c t 2 p r o f e s s o r o u t o f 10 a n d 3 l e c t u r e r s o u t o f 20 \\ ∴ N u m b e r o f w a y s o f s e l e c t i o n = {}_{10}{C_{2}} \times {}_{20}{C_{3}} \\ (b) W h e n a p a r t i c u l a r p r o f e s s o r i s i n c l u d e d t a k e n n u m b e r o f w a y s = {}_{10 - 1}{C_{1}} \times {}_{20}{C_{3}} = {}_{9}{C_{1}} \times {}_{20}{C_{3}} \\ (c) W h e n a p a r t i c u l a r l e c t u r e r i s i n c l u d e d, t h e n n u m b e r o f w a y s = {}_{10}{C_{2}} \times {}_{19}{C_{2}} \\ (d) W h e n a p a r t i c u l a r l e c t u r e r i s e x c l u d e d, t h e n n u m b e r o f w a y s = {}_{10}{C_{2}} \times {}_{19}{C_{3}} \\ H e n c e, t h e r e q u i r e d m a t c h i n g i s \\ (a) \leftrightarrow (i v), (b) \leftrightarrow (i i i), (c) \leftrightarrow (i i) a n d (d) \leftrightarrow (i) \end{array}$

How many words (with or without dictionary meaning) can be made from the letters of the word MONDAY, assuming that no letter is repeated, if

C₁ | C₂

(a) 4 letters are used at a time | (i) 720

(b) All letters are used at a time | (ii) 240

(c) All letters are used but the first is a vowel | (iii) 360

This is a matching answer type question as classified in NCERT Exemplar

$\begin{array}{l} (a) 4 l e t t e r s a r e u s e d a t a t i m e = {}_{6}{P_{4}} = \frac{6!}{2!} = 360 \\ (b) A l l l e t t e r s a r e u s e d a t a t i m e = {}_{6}{P_{6}} = 6! = 720 \\ (c) A l l l e t t e r s a r e u s e d b u t f i r s t l e t t e r i s v o w e l = 2 \times 5! = 2 \times 120 = 240 \\ H e n c e, t h e r e q u i r e d m a t c h i n g i s \\ (a) \leftrightarrow (i i i), (b) \leftrightarrow (i), (c) \leftrightarrow (i i) \end{array}$

There are 3 books on Mathematics, 4 on Physics, and 5 on English. How many different collections can be made such that each collection consists of:

C₁ | C₂

(a) One book of each subject; | (i) 3968

(b) At least one book of each subject: | (ii) 60

(c) At least one book of English: | (iii) 3255

This is a matching answer type question as classified in NCERT Exemplar

$\begin{array}{l} W e h a v e 3 b o o k s o f M a t h e m a t i c s, 4 o f P h y s i c s a n d 5 o n E n g l i s h \\ (a) O n e b o o k o f e a c h s u b j e c t = {}_{3}{C_{1}} \times {}_{4}{C_{1}} \times {}_{5}{C_{1}} = 3 \times 4 \times 5 = 60 \\ (b) A t l e a s t o n e b o o k o f e a c h s u b j e c t = (2^{3} - 1) \times (2^{4} - 1) \times (2^{5} - 1) = 7 \times 15 \times 31 = 3255 \\ (c) A t l e a s t o n e b o o k o f E n g l i s h = (2^{5} - 1) \times 2^{7} = 31 \times 128 = 3968 \\ H e n c e, t h e r e q u i r e d m a t c h i n g i s \\ (a) \leftrightarrow (i i), (b) \leftrightarrow (i i i) a n d (c) \leftrightarrow (i) \end{array}$

Using the digits 1, 2, 3, 4, 5, 6, 7, a number of 4 different digits is formed. Find:

C₁ C₂

(a) How many numbers are formed? | (i) 840

(b) How many numbers are exactly divisible by 2? | (ii) 200

(c) How many numbers are exactly divisible by 25? | (iii) 360

(d) How many of these are exactly divisible by 4? | (iv) 40

This is a matching answer type question as classified in NCERT Exemplar

$\begin{array}{l} (a) T o t a l o f 4 d i g i t n u m b e r f o r m e d w i t h 1, 2, 3, 4, 5, 6, 7 = {}_{7}{P_{4}} = \frac{7!}{(7 - 4)!} \\ = \frac{7 \times 6 \times 5 \times 4 \times 3!}{3!} = 840 \\ (b) W h e n a n u m b e r i s d i v i s i b l e b y 2 = 4 \times 5 \times 6 \times 3 = 360 \\ (c) T o t a l n u m b e r s w h i c h a r e d i v i s i b l e b y 25 = 40 \\ (d) T o t a l n u m b e r s w h i c h a r e d i v i s i b l e b y 4 (l a s t t w o d i g i t s i s d i v i s i b l e b y 4) = 200 \\ H e n c e, t h e r e q u i r e d m a t c h i n g i s \\ (a) \leftrightarrow (i), (b) \leftrightarrow (i i i), (c) \leftrightarrow (i v) a n d (d) \leftrightarrow (i i) \end{array}$

Maths NCERT Exemplar Solutions Class 11th Chapter Seven Exam

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