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New answer posted

a year ago

Class 11th Exemplar Solution

Find the number of positive integers greater than 6000 and less than 7000 which are divisible by 5, provided that no digit is to be repeated.

0 Follower 8 Views

Payal Gupta

Contributor-Level 10

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

$\begin{array}{l} A n y n u m b e r d i v i s i b l e b y 5, i t s u n i t p l a c e m u s t h a v e 0 o r 5 \\ W e h a v e t o f i n d 4 - d i g i t n u m b e r g r e a t e r t h a n 6000 a n d l e s s t h a n 7000 . \\ S o, t h e u n i t p l a c e c a n b e f i l l e d w i t h 2 w a y s (0 o r 5) since, r e p e t i t i o n i s n o t a l l o w e d . \\ ∴ t e n s p l a c e c a n b e f i l l e d w i t h 7 w a y s a n d h u n d r e d s p l a c e c a n b e f i l l e d w i t h 8 w a y s . \\ B u t t h e r e q u i r e d n u m b e r i s g r e a t e r t h a n 6000 a n d l e s s t h a n 7000 . \\ S o, t h o u s a n d p l a c e c a n b e f i l l e d w i t h 1 d i g i t s i . e . 6 \\ \underset{1}{T h} \underset{8}{H} \underset{7}{T} \underset{2}{O} \\ S o, t h e t o t a l n u m b e r o f integers = 1 * 8 * 7 * 2 = 112 \\ H e n c e, t h e r e q u i r e d n u m b e r o f integers = 112 \end{array}$

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New answer posted

a year ago

Class 11th Exemplar Solution

Find the coefficient of $x$ in the expansion of

$(1 - 3 x + 7 x^{2}) {(1 - x)}^{16} .$

0 Follower 17 Views

Vishal Baghel

Contributor-Level 10

This is a Short Answer Type Questions as classified in NCERT Exemplar
Sol:

$\begin{array}{l} T h e g i v e n expression i s (1 - 3 x + 7 x^{2}) {(1 - x)}^{16} \\ = (1 - 3 x + 7 x^{2}) [{}_{16}{C_{0}} {(1)}^{16} {(- x)}^{0} + {}_{16}{C_{1}} {(1)}^{15} (- x) + {}_{16}{C_{2}} {(1)}^{14} {(- x)}^{2} + \dots] \\ = (1 - 3 x + 7 x^{2}) (1 - 16 x + 120 x^{2} + \dots) \\ C o l l e c t i n g t h e t e r m c o n t a i n i n g x, w e g e t - 16 x - 3 x = - 19 x \\ H e n c e, t h e c o e f f i c i e n t o f x = - 19 \end{array}$

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New answer posted

a year ago

Class 11th Exemplar Solution

If the term free from $x$ in the expansion of x – k/x² ¹⁰ is 405, find the value of $k$ .

0 Follower 4 Views

Vishal Baghel

Contributor-Level 10

This is a Short Answer Type Questions as classified in NCERT Exemplar
Sol:

$\begin{array}{l} T h e g i v e n expression i s {(\sqrt[]{x} - \frac{K}{x^{2}})}^{10} \\ G e n e r a l T e r m T_{r + 1} = {}_{n}{C_{r}} x^{n - r} y^{r} \\ = {}_{10}{C_{r}} {(\sqrt[]{x})}^{10 - r} {(\frac{- K}{x^{2}})}^{r} = {}_{10}{C_{r}} {(x)}^{\frac{10 - r}{2}} {(- K)}^{r} (\frac{1}{x^{2 r}}) \\ = {}_{10}{C_{r}} {(x)}^{\frac{10 - r}{2} - 2 r} {(- K)}^{r} = {}_{10}{C_{r}} {(x)}^{\frac{10 - r - 4 r}{2}} {(- K)}^{r} \\ = {}_{10}{C_{r}} {(x)}^{\frac{10 - 5 r}{2}} {(- K)}^{r} \\ f o r g e t t i n g t h e t e r m i n d e p e n d e n t o f x, \\ \frac{10 - 5 r}{2} = 0 \Rightarrow r = 2 \\ O n p u t t i n g t h e v a l u e o f r i n t h e a b o v e expression, w e g e t \\ = {}_{10}{C_{2}} {(- K)}^{2} \\ A c c o r d i n g t o t h e c o n d i t i o n o f t h e q u e s t i o n, w e h a v e \\ {}_{10}{C_{2}} K^{2} = 405 \Rightarrow \frac{10.9}{2.1} K^{2} = 405 \\ \Rightarrow 45 K^{2} = 405 \Rightarrow K^{2} = \frac{405}{45} = 9 \\ ∴ K = \pm 3 \\ H e n c e, t h e v a l u e o f K = \pm 3 \end{array}$

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New answer posted

a year ago

Class 11th Exemplar Solution

Find the number of different words that can be formed from the letters of the word 'TRIANGLE' so that no vowels are together.

0 Follower 11 Views

Payal Gupta

Contributor-Level 10

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

$\begin{array}{l} T o t a l n u m b e r o f w o r d s i n' T R I A N G L E' = 8 \\ O u t o f 5 a r e c o n s o n a n t s a n d 3 a r e v o w e l s \\ I f v o w e l s a r e n o t t o g e t h e r, t a k e n w e h a v e t h e f o l l o w i n g a r r a n g e m e n t \\ V Unsupported ? C Unsupported ? V Unsupported ? C Unsupported ? V Unsupported ? C Unsupported ? V Unsupported ? C Unsupported ? V Unsupported ? C Unsupported ? V \\ c o n s o n a n t c a n b e a r r a n g e d i n 5! = 120 w a y s \\ V o w e l o c c u p y 6 p l a c e s \\ ∴ 3 v o w e l s c a n b e a r r a n g e d i n 6 p l a c e s = {}_{6}{P_{3}} \\ = \frac{6!}{(6 - 3)!} = \frac{6!}{3!} = 120 w a y s \\ S o, t h e t o t a l a r r a n g e m e n t = 120 * 120 = 14400 w a y s \\ H e r e, t h e r e q u i r e d a r r a n g e m e n t = 14400 w a y s . \end{array}$

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New answer posted

a year ago

Class 11th Exemplar Solution

Find the number of permutations of n distinct things taken r together, in which 3 particular things must occur together.

0 Follower 3 Views

Payal Gupta

Contributor-Level 10

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

$\begin{array}{l} T o t a l n u m b e r o f t h i n g s = n \\ 3 t h i n g s m u s t b e t o g e t h e r \\ ∴ t h e n u m b e r o f r e m a i n i n g t h i n g s = n - 3 \\ N u m b e r o f t h i n g s t o b e s e l e c t e d = r \\ O u t o f r, 3 a r e a l w a y s t o g e t h e r \\ ∴ N u m b e r o f w a y s o f s e l e c t i o n = {}_{n - 3}{C_{r - 2}} \\ N o w p e r m u t a t i o n o f 3 t h i n g s w h i c h a r e a l w a y s t o g e t h e r = 3! \\ N u m b e r o f p e r m u t a t i o n o f (r - 2) t h i n g s = (r - 2)! \\ ∴ T o t a l n u m b e r o f a r r a n g e m e n t s = {}_{n - 3}{C_{r - 2}} * (r - 2)! * 3! \\ H e n c e, t h e r e q u i r e d a r r a n g e m e n t s = {}_{n - 3}{C_{r - 2}} * (r - 2)! * 3! . \end{array}$

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New answer posted

a year ago

Class 11th Exemplar Solution

Find the term independent of $x$ , $x \neq 0$ , in the expansion of (3x²/2 – 1/3x)¹⁵

0 Follower 6 Views

Vishal Baghel

Contributor-Level 10

This is a Short Answer Type Questions as classified in NCERT Exemplar
Sol:

$\begin{array}{l} G e n e r a l T e r m T_{r + 1} = {}_{n}{C_{r}} x^{n - r} y^{r} \\ = {}_{15}{C_{r}} {(\frac{3 x^{2}}{2})}^{15 - r} {(- \frac{1}{3 x})}^{r} = {}_{15}{C_{r}} {(\frac{3}{2})}^{15 - r} {(x)}^{30 - 2 r} {(- \frac{1}{3})}^{r} \frac{1}{x^{r}} \\ = {}_{15}{C_{r}} {(\frac{3}{2})}^{15 - r} {(x)}^{30 - 2 r - r} {(- 1)}^{r} \frac{1}{{(3)}^{r}} \\ = {}_{15}{C_{r}} {(\frac{3}{2})}^{15 - r} {(x)}^{30 - 3 r} {(- 1)}^{r} \frac{1}{{(3)}^{r}} \\ f o r g e t t i n g t h e t e r m i n d e p e n d e n t o f x, \\ 30 - 3 r = 0 \Rightarrow r = 10 \\ O n p u t t i n g t h e v a l u e o f r i n t h e a b o v e expression, w e g e t \\ = {}_{15}{C_{10}} {(\frac{3}{2})}^{15 - 10} {(- 1)}^{10} \frac{1}{{(3)}^{10}} = {}_{15}{C_{10}} \frac{{(3)}^{5}}{{(2)}^{5}} . \frac{1}{{(3)}^{10}} \\ = {}_{15}{C_{10}} . \frac{1}{{(2)}^{5} . {(3)}^{5}} = {}_{15}{C_{10}} {(\frac{1}{6})}^{5} \\ H e n c e, t h e r e q u i r e d t e r m = {}_{15}{C_{10}} {(\frac{1}{6})}^{5} \end{array}$

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New answer posted

a year ago

Class 11th Exemplar Solution

A bag contains 5 black and 6 red balls. Determine the number of ways in which 2 black and 3 red balls can be selected from the lot.

0 Follower 3 Views

Payal Gupta

Contributor-Level 10

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

$\begin{array}{l} G i v e n t h a t b a g c o n t a i n s 5 b l a c k a n d 6 r e d b a l l s . \\ N u m b e r o f w a y s o f s e l e c t i n g 2 b l a c k b a l l s o u t o f 5 b l a c k b a l l s = {}_{5}{C_{2}} \\ a n d n u m b e r o f w a y s o f s e l e c t i n g 3 r e d b a l l s o u t o f 6 r e d b a l l s = {}_{6}{C_{3}} \\ ∴ T o t a l n u m b e r o f w a y s o f s e l e c t i n g 2 b l a c k a n d 3 r e d b a l l s = {}_{5}{C_{2}} * {}_{6}{C_{3}} \\ = \frac{5.4}{2.1} * \frac{6.5.4}{3.2.1} = 10 * 20 = 200 w a y s \\ H e n c e, t h e r e q u i r e d w a y s o f s e l e c t i n g t h e b a l l s = 200 . \end{array}$

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New answer posted

a year ago

Class 11th Exemplar Solution

How many automobile license plates can be made if each plate contains two different letters followed by three different digits?

0 Follower 7 Views

Payal Gupta

Contributor-Level 10

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

$\begin{array}{l} T o t a l n u m b e r o f p e r s o n s = 12 \\ N u m b e r o f p e r s o n s t o b e s e l e c t e d = 5 \\ W e h a v e 26 E n g l i s h a l p h a b e t a n d 10 d i g i t s (0 t o 9) \\ Since, i t i s g i v e n t h a t e a c h p l a t e c o n t a i n s 2 d i f f e r e n t l e t t e r s f o l l o w e d b y 3 d i f f e r e n t d i g i t s . \\ ∴ N u m b e r o f a r r a n g e m e n t o f 26 l e t t e r s t a k e n 2 a t a t i m e = {}_{26}{P_{2}} = \frac{26!}{(26 - 2)!} = \frac{26!}{24!} \\ = \frac{26.25.24!}{24!} = 650 \\ T h r e e d i g i t n u m b e r c a n b e f o r m e d o u t o f 10 d i g i t = {}_{10}{P_{3}} = \frac{10!}{7!} \\ = \frac{10.9.8.7!}{7!} = 720 \\ ∴ T o t a s l n u m b e r o f l i c e n s e p l a t e s = 650 * 720 = 468000 \\ H e n c e, t h e r e q u i r e d n u m b e r o f p l a t e s = 468000 . \end{array}$

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New answer posted

a year ago

Class 11th Exemplar Solution

How many committees of five persons with a chairperson can be selected from 12 persons?

0 Follower 2 Views

Payal Gupta

Contributor-Level 10

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

$\begin{array}{l} T o t a l n u m b e r o f p e r s o n s = 12 \\ N u m b e r o f p e r s o n s t o b e s e l e c t e d = 5 \\ O u t o f 5, t h e r e i s a c h a i r p e r s o n \\ ∴ N u m b e r o f w a y s o f s e l e c t i n g a c h a i r p e r s o n = {}_{12}{C_{1}} = 12 \\ N u m b e r o f w a y s o f s e l e c t i n g o t h e r 4 n u m b e r s o u t o f r e m a i n i n g 11 p e r s o n s = {}_{11}{C_{4}} \\ ∴ T o t a l n u m b e r o f w a y s = {}_{12}{C_{1}} * {}_{11}{C_{4}} \\ = 12 * \frac{11.10.9.8}{4.3.2.1} = 12 * 330 = 3960 \\ H e n c e, t h e r e q u i r e d n u m b e r o f w a y s = 3960 . \end{array}$

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New answer posted

a year ago

Class 11th Exemplar Solution

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?

0 Follower 2 Views

Payal Gupta

Contributor-Level 10

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

$\begin{array}{l} T o t a l n u m b e r o f p e r s o n s = 8 \\ N u m b e r o f p e r s o n s t o b e s e l e c t e d = 6 \\ C o n d i t i o n i s t h a t i f A i s c h o o s e n, B m u s t b e c h o o s e n \\ C a s e I : W h e n A i s c h o o s e n, B m u s t b e c h o o s e n \\ N u m b e r o f w a y s = {}_{6}{C_{4}} [? A a n d B a r e s e t t o b e c h o o s e n] \\ C a s e I I : W h e n A i s n o t c h o o s e n, t h e n B m a y b e c h o o s e n \\ N u m b e r o f w a y s = {}_{7}{C_{6}} \\ S o, t h e t o t a l n u m b e r o f w a y s = {}_{6}{C_{4}} + {}_{7}{C_{6}} [? T h e r e a r e t w o c a s e s] \\ = {}_{6}{C_{2}} + {}_{7}{C_{1}} [? {}_{n}{C_{r}} = {}_{n}{C_{n - r}}] \\ = \frac{6.5}{2.1} + 7 = 15 + 7 = 22 w a y s \\ H e n c e, t h e r e q u i r e d n u m b e r o f w a y s = 22 . \end{array}$

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