Class 11th

Contributor-Level 10

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

Sol:

$\begin{array}{l} L e t u s c o n s i d e r a G . P, a, a r a n d a r^{2} \\ I f i t i s i n A . P t h e n a r - a \neq a r^{2} - a r \\ H e n c e, t h e g i v e n s t a t e m e n t i s T r u e . \end{array}$

0 0

New answer posted

a year ago

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

0 Follower 4 Views

Contributor-Level 10

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

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

0 0

New answer posted

a year ago

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

0 Follower 2 Views

Contributor-Level 10

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

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

0 0

New answer posted

a year ago

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.

0 Follower 7 Views

Class 11th Maths NCERT Exemplar Solutions Class 11th Chapter Nine

Contributor-Level 10

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

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

0 0

New answer posted

a year ago

The sum of terms equidistant from the beginning and end in an A.P. is equal to . .

0 Follower 4 Views

Class 11th Maths NCERT Exemplar Solutions Class 11th Chapter Nine

Contributor-Level 10

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

Sol:

$\begin{array}{l} L e t A . P b e a, a + d, a + 2 d, a + 3 d, \dots, a + (n - 1) d \\ T a k i n g f i r s t a n d l a s t t e r m \\ a_{1} + a_{n} = a + a + (n - 1) d = 2 a + (n - 1) d \\ T a k i n g second a n d second l a s t t e r m \\ a_{2} + a_{n - 1} = (a + d) + [a + (n - 2) d] = 2 a + (n - 1) d = a_{1} + a_{n} \\ T a k i n g t h i r d f r o m t h e b e g i n n i n g a n d t h i r d f r o m t h e e n d \\ a_{3} + a_{n - 2} = (a + 2 d) + [a + (n - 3) d] = 2 a + (n - 1) d = a_{1} + a_{n} \\ F r o m t h e a b o v e p a t t e r n, w e o b s e r v e t h a t t h e s u m o f t e r m s equidistant f r o m t h e b e g i n n i n g \\ a n d t h e e n d i n a n A . P i s e q u a l t o t h e [f i r s t t e r m + l a s t t e r m] \\ H e n c e, t h e c o r r e c t v a l u e o f t h e f i l l e r i s [f i r s t t e r m + l a s t t e r m] . \end{array}$

0 0

New answer posted

a year ago

For $a, b, c$ to be in G.P., the value of $\frac{a - b}{b - c}$ is equal to . .

0 Follower 6 Views

Contributor-Level 10

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

Sol:

$\begin{array}{l} Since a, b a n d c a r e i n G . P \\ ∴ \frac{b}{a} = \frac{c}{b} = r (constant) \\ \Rightarrow b = a r a n d c = b r \Rightarrow c = a r . r = a r^{2} \\ S o, \frac{a - b}{b - c} = \frac{a - a r}{a r - a r^{2}} = \frac{a (1 - r)}{a r (1 - r)} = \frac{1}{r} = \frac{a}{b} = \frac{b}{c} \\ H e n c e, t h e c o r r e c t v a l u e o f t h e f i l l e r i s \frac{a}{b} o r \frac{b}{c} . \end{array}$

0 0

New answer posted

a year ago

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 ______.

0 Follower 2 Views

Contributor-Level 10

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

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

0 0

New answer posted

a year ago

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.]

0 Follower 14 Views

Contributor-Level 10

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

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

0 0

New answer posted

a year ago

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 ______.

0 Follower 2 Views

Class 11th Maths NCERT Exemplar Solutions Class 11th Chapter Nine

Contributor-Level 10

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

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

0 0

New answer posted

a year ago

The lengths of three unequal edges of a rectangular solid block are in G.P. The volume of the block is $216 c m^{3}$ and the total surface area $252 c m^{2}$ is . The length of the longest edge is:

(a) 12

(b) 6

(c) 18

(d) 3

0 Follower 3 Views