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Exemplar Solution Class 11th

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

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Payal Gupta | Contributor-Level 10

a year ago

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

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

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Exemplar Solution Class 11th

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

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Asked by Payal Gupta

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Answered by

Payal Gupta | Contributor-Level 10

a year ago

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

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

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Exemplar Solution Class 11th

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!

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Asked by Payal Gupta

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Answered by

Payal Gupta | Contributor-Level 10

a year ago

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

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

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Exemplar Solution Class 11th

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

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Asked by Payal Gupta

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Answered by

Payal Gupta | Contributor-Level 10

a year ago

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

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

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Exemplar Solution Class 11th

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.

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Answered by

Payal Gupta | Contributor-Level 10

a year ago

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

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

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Exemplar Solution Class 11th

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.

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Payal Gupta | Contributor-Level 10

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

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Exemplar Solution Maths NCERT Exemplar Solutions Class 11th Chapter Nine

If the sum of $n$ terms of a sequence is a quadratic expression, then it always represents an A.P.

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Asked by Alok Kumar Singh

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Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

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

Sol:

$\begin{array}{l} L e t S_{n} = a n^{2} + b n + c [Q u a d r a t i c expression] \\ S_{1} = a + b + c \\ ∴ a_{1} = a + b + c \\ S_{2} = 4 a + 2 b + c \\ a_{2} = S_{2} - S_{1} = (4 a + 2 b + c) - (a + b + c) = 3 a + b \\ S_{3} = 9 a + 3 b + c \\ a_{3} = S_{3} - S_{2} = (9 a + 3 b + c) - (4 a + 2 b + c) = 5 a + b \\ C o m m o n d i f f e r e n c e d = a_{2} - a_{1} = (3 a + b) - (a + b + c) = 2 a - c \\ a n d d = a_{3} - a_{2} = (5 a + b) - (3 a + b) = 2 a \\ H e r e, w e o b s e r v e t h a t a_{2} - a_{1} \neq a_{3} - a_{2} \\ S o i t d o e s n o t r e p r e s e n t a n A . P \\ H e n c e, t h e g i v e n s t a t e m e n t i s F a l s e . \end{array}$

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Exemplar Solution Class 11th

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 $\frac{11! (9!) (9!)}{5! 6!}$ .

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Asked by Payal Gupta

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Answered by

Payal Gupta | Contributor-Level 10

a year ago

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

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

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Exemplar Solution Maths NCERT Exemplar Solutions Class 11th Chapter Nine

The sum or difference of two G.P.s is again a G.P.

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Alok Kumar Singh | Contributor-Level 10

a year ago

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

Sol:

$\begin{array}{l} L e t u s c o n s i d e r t w o G . P .' s \\ a_{1}, a_{1} r_{1}, a_{1} r_{1}^{2}, a_{1} r_{1}^{3} \dots a_{1} r_{1}^{n - 1} \\ a n d a_{2}, a_{2} r_{2}, a_{2} r_{2}^{2}, a_{2} r_{2}^{3} \dots a_{2} r_{2}^{n - 1} \\ N o w S u m o f t w o G . P s \\ (a_{1} + a_{2}) + (a_{1} r_{1} + a_{2} r_{2}) + (a_{1} r_{1}^{2} + a_{2} r_{2}^{2}) + \dots \\ N o w \frac{T_{2}}{T_{1}} = \frac{a_{1} r_{1} + a_{2} r_{2}}{a_{1} + a_{2}} a n d \frac{T_{3}}{T_{2}} = \frac{a_{1} r_{1}^{2} + a_{2} r_{2}^{2}}{a_{1} r_{1} + a_{2} r_{2}} \\ B u t \frac{a_{1} r_{1} + a_{2} r_{2}}{a_{1} + a_{2}} \neq \frac{a_{1} r_{1}^{2} + a_{2} r_{2}^{2}}{a_{1} r_{1} + a_{2} r_{2}} \\ L e t u s c o n s i d e r t h e d i f f e r e n c e G . P' s \\ (a_{1} - a_{2}) + (a_{1} r_{1} - a_{2} r_{2}) + (a_{1} r_{1}^{2} - a_{2} r_{2}^{2}) + \dots \\ N o w \frac{T_{2}}{T_{1}} = \frac{a_{1} r_{1} - a_{2} r_{2}}{a_{1} - a_{2}} a n d \frac{T_{3}}{T_{2}} = \frac{a_{1} r_{1}^{2} - a_{2} r_{2}^{2}}{a_{1} r_{1} - a_{2} r_{2}} \\ B u t \frac{T_{2}}{T_{1}} \neq \frac{T_{3}}{T_{2}} \\ H e n c e, t h e g i v e n s t a t e m e n t i s F a l s e . \end{array}$

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Exemplar Solution Class 11th

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.

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Asked by Payal Gupta

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Answered by

Payal Gupta | Contributor-Level 10

a year ago

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

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

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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

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.

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.

If the sum of $n$ terms of a sequence is a quadratic expression, then it always represents an A.P.

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 $\frac{11! (9!) (9!)}{5! 6!}$ .

The sum or difference of two G.P.s is again a G.P.

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.