Class 11th

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

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Contributor-Level 10

This is an Objective Type Questions as classified in NCERT Exemplar

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

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New answer posted

a year ago

The expression $7^{9} + 9^{7}$ is divisible by 64.

[Hint: $7^{9} + 9^{7} = {(1 + 8)}^{9} - {(1 - 8)}^{7}$ .]

0 Follower 7 Views

Contributor-Level 10

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

$\begin{array}{l} 7^{9} + 9^{7} = {(1 + 8)}^{7} - {(1 - 8)}^{9} \\ = [{}_{7}{C_{0}} + {}_{7}{C_{1}} . 8 + {}_{7}{C_{2}} {(8)}^{2} + {}_{7}{C_{3}} {(8)}^{3} + \dots + {}_{7}{C_{7}} {(8)}^{7}] - [\begin{array}{l} {}_{9}{C_{0}} - {}_{9}{C_{1}} . 8 + {}_{9}{C_{2}} {(8)}^{2} - {}_{9}{C_{3}} {(8)}^{3} \\ + \dots {}_{9}{C_{9}} {(8)}^{9} \end{array}] \\ = (7 * 8 + 9 * 8) + (21 * 8^{2} - 36 * 8^{2}) + \dots \\ = (56 + 72) + (21 - 36) * 8^{2} + \dots \\ = 128 + 64 (21 - 36) + \dots \\ = 64 [2 + (21 - 36) + \dots] \\ w h i c h i s d i v i s i b l e b y 64 . \\ H e n c e, t h e g i v e n s t a t e m e n t i s' T r u e' . \end{array}$

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New answer posted

a year ago

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

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

Contributor-Level 10

This is an Objective Type Questions as classified in NCERT Exemplar

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

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New answer posted

a year ago

If $θ_{1}, θ_{2}, θ_{3}, . . ., θ_{n}$ are in A.P., whose common difference is $d$ , show that

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alok kumar singh

Contributor-Level 10

This is a Long Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} Since θ_{1}, θ_{2}, θ_{3}, \dots, θ_{n} a r e i n A . P . \\ ∴ θ_{2} - θ_{1} = θ_{3} - θ_{2} = \dots = θ_{n} - θ_{n - 1} = d \\ N o w w e h a v e t o p r o v e t h a t \\ s e c θ_{1} . s e c θ_{2} + s e c θ_{2} . s e c θ_{3} + \dots + s e c θ_{n - 1} . s e c θ_{n} = \frac{t a n θ_{n} - t a n θ_{1}}{s i n d} \\ L H S . \frac{s i n d}{s i n d} [s e c θ_{1} . s e c θ_{2} + s e c θ_{2} . s e c θ_{3} + \dots + s e c θ_{n - 1} . s e c θ_{n}] \\ T a k i n g o n l y \frac{s i n d [s e c θ_{1} . s e c θ_{2}]}{s i n d} = \frac{s i n d [\frac{1}{c o s θ_{1}} . \frac{1}{c o s θ_{2}}]}{s i n d} \\ = \frac{s i n (θ_{2} - θ_{1})}{s i n d} . \frac{1}{c o s θ_{1} c o s θ_{2}} \\ = \frac{1}{s i n d} [\frac{s i n θ_{2} c o s θ_{1} - c o s θ_{2} s i n θ_{1}}{c o s θ_{1} c o s θ_{2}}] \\ = \frac{1}{s i n d} [\frac{s i n θ_{2} c o s θ_{1}}{c o s θ_{1} c o s θ_{2}} - \frac{c o s θ_{2} s i n θ_{1}}{c o s θ_{1} c o s θ_{2}}] \\ = \frac{1}{s i n d} [t a n θ_{2} - t a n θ_{1}] \\ S i m i l a r l y, w e c a n s o l v e o t h e r t e r m s w h i c h w i l l b e \\ \frac{1}{s i n d} [t a n θ_{3} - t a n θ_{2}] a n d \frac{1}{s i n d} [t a n θ_{4} - t a n θ_{3}] \\ H e r e L H S = \frac{1}{s i n d} [t a n θ_{2} - t a n θ_{1} + t a n θ_{3} - t a n θ_{2} + \dots + t a n θ_{n} - t a n θ_{n - 1}] \\ = \frac{1}{s i n d} [- t a n θ_{1} + t a n θ_{n}] = \frac{t a n θ_{n} - t a n θ_{1}}{s i n d} R H S . \\ L H S = R H S H e n c e p r o v e d . \end{array}$

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New answer posted

a year ago

The sum of the series ^{10 20} _{r =0} C_ris 2¹⁹ + ²⁰ C₁₀/2

0 Follower 14 Views

Contributor-Level 10

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

$\begin{array}{l} \sum_{r = 0}^{10} {}_{20}{C_{r}} = {}_{20}{C_{0}} + {}_{20}{C_{1}} + {}_{20}{C_{2}} + {}_{20}{C_{3}} + \dots + {}_{20}{C_{10}} \\ = {}_{20}{C_{0}} + {}_{20}{C_{1}} + \dots + {}_{20}{C_{10}} + {}_{20}{C_{11}} + \dots + {}_{20}{C_{20}} - ({}_{20}{C_{11}} + \dots + {}_{20}{C_{20}}) \\ = 2^{20} - ({}_{20}{C_{11}} + \dots + {}_{20}{C_{20}}) \\ H e n c e, t h e g i v e n s t a t e m e n t i s' F a l s e' \end{array}$

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New answer posted

a year ago

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

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

Contributor-Level 10

This is an Objective Type Questions as classified in NCERT Exemplar

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

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New answer posted

a year ago

If $A$ is the arithmetic mean and $G_{1}, G_{2}$ be two geometric means between any two numbers, then prove that

$2 A = \frac{G_{1}^{2}}{G_{2}} + \frac{G_{2}^{2}}{G_{1}}$

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alok kumar singh

Contributor-Level 10

This is a Long Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} L e t t h e t w o n u m b e r s b e x a n d y \\ ∴ A = \frac{x + y}{2} \dots (i) \\ I f G_{1} a n d G_{2} b e t h e g e o m e t r i c m e a n s b e t w e e n x a n d y t h e n x, G_{1}, G_{2}, y a r e i n G . P . \\ t h e n y = x r^{4 - 1} [? a_{n} = a r^{n - 1}] \\ \Rightarrow y = x r^{3} \Rightarrow \frac{y}{x} = r^{3} \\ \Rightarrow r = {(\frac{y}{x})}^{1 / 3} \\ N o w G_{1} = x r = x {(\frac{y}{x})}^{1 / 3} [? r = {(\frac{y}{x})}^{1 / 3}] \\ a n d G_{2} = x r^{2} = x {(\frac{y}{x})}^{2 / 3} \\ ∴ F r o m R H S \frac{G_{1}^{2}}{G_{2}} + \frac{G_{2}^{2}}{G_{1}} = \frac{x^{2} {(\frac{y}{x})}^{2 / 3}}{x {(\frac{y}{x})}^{2 / 3}} + \frac{x^{2} {(\frac{y}{x})}^{4 / 3}}{x {(\frac{y}{x})}^{1 / 3}} \\ = x + x {(\frac{y}{x})}^{\frac{4}{3} - \frac{1}{3}} = x + x (\frac{y}{x}) = x + y = 2 A L H S [using e q . (i)] \\ L H S = R H S H e n c e p r o v e d . \end{array}$

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New answer posted

a year ago

If 2515 is divided by 13, the remainder is _________________ .

0 Follower 3 Views

Contributor-Level 10

This is a Fill in the Blanks Type Questions as classified in NCERT Exemplar
Sol:

$\begin{array}{l} L e t 2 5^{15} = {(26 - 1)}^{15} \\ = {}_{15}{C_{0}} {(26)}^{15} {(- 1)}^{0} + {}_{15}{C_{1}} {(26)}^{14} {(- 1)}^{1} + {}_{15}{C_{2}} {(26)}^{13} {(- 1)}^{2} + \dots + {}_{15}{C_{15}} {(- 1)}^{15} \\ = 2 6^{15} - 15 {(26)}^{14} + \dots - 1 - 13 + 13 \\ = 2 6^{15} - 15 {(26)}^{14} + \dots - 13 + 12 \\ = 13 λ + 12 \\ ∴ T h e r e m a i n d e r = 12 \\ H e n c e, t h e v a l u e o f t h e f i l l e r i s 12 . \end{array}$

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New answer posted

a year ago

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

0 Follower 13 Views

Contributor-Level 10

This is an Objective Type Questions as classified in NCERT Exemplar

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

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New answer posted

a year ago

The position of the term independent of $x$ in the expansion of $\sqrt {\frac{x}{3} + \frac{3}{{2 x}^{2}}}^{10}$ is _________________ .

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