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Exemplar Solution Class 11th

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.

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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 long answer type question as classified in NCERT Exemplar

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

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Exemplar Solution Class 11th

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

8 Views|Posted a year ago

Asked by Payal Gupta

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Answered by

Payal Gupta | Contributor-Level 10

a year ago

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

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

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Exemplar Solution Class 11th

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

3 Views|Posted a year ago

Asked by Payal Gupta

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Answered by

Payal Gupta | Contributor-Level 10

a year ago

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

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

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Exemplar Solution Class 11th

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?

2 Views|Posted a year ago

Asked by Payal Gupta

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Answered by

Payal Gupta | Contributor-Level 10

a year ago

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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Exemplar Solution Class 11th

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.]

8 Views|Posted a year ago

Asked by Payal Gupta

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Answered by

Payal Gupta | Contributor-Level 10

a year ago

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

$\begin{array}{l} T o t a l n u m b e r o f points = 12 \\ O u t o f 18 n u m b e r s, 5 a r e c o l l i n e a r a n d w e g e t a s t r a i g h t l i n e b y j o i n i n g a n y t w o points. \\ ∴ T o t a l n u m b e r o f s t r a i g h t l i n e f o r m e d b y j o i n i n g 2 points o u t o f 18 points = {}_{18}{C_{2}} \\ N u m b e r o f s t r a i g h t l i n e s f o r m e d b y j o i n i n g 2 points o u t o f 5 points = {}_{5}{C_{2}} \\ B u t 5 points a r e c o l l i n e a r a n d w e g e t o n l y o n e l i n e w h e n t h e y a r e j o i n e d p a i r w i s e . \\ S o, t h e r e q u i r e d n u m b e r o f s t r a i g h t l i n e s a r e \\ = {}_{18}{C_{2}} - {}_{5}{C_{2}} + 1 = \frac{18.17}{2.1} - \frac{5.4}{2.1} + 1 = 153 - 10 + 1 = 144 \\ H e n c e, t o t a l n u m b e r o f s t r a i g h t l i n e s = 144 . \end{array}$

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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. Find the number of different ways of doing questions.

4 Views|Posted a year ago

Asked by Payal Gupta

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Answered by

Payal Gupta | Contributor-Level 10

a year ago

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

$\begin{array}{l} T o t a l n u m b e r o f q u e s t i o n s = 12 \\ N u m b e r o f q u e s t i o n s i n e a c h g r o u p = 6 \\ 7 q u e s t i o n s a r e t o b e a t t e m p t e d b u t n o t m o r e t h a n 5 q u e s t i o n s f r o m e i t h e r g r o u p \\ ∴ T o t a l n u m b e r o f w a y s = {}_{6}{C_{5}} \times {}_{6}{C_{2}} + {}_{6}{C_{4}} \times {}_{6}{C_{3}} + {}_{6}{C_{3}} \times {}_{6}{C_{4}} + {}_{6}{C_{2}} \times {}_{6}{C_{5}} \\ = 2 [{}_{6}{C_{5}} \times {}_{6}{C_{2}} + {}_{6}{C_{4}} \times {}_{6}{C_{3}}] = 2 [6 \times \frac{6.5}{2.1} + \frac{6.5.4.3}{4.3.2.1} \times \frac{6.5.4}{3.2.1}] \\ = 2 [6 \times 15 + 15 \times 20] = 2 [90 + 300] = 2 \times 390 = 780 \\ H e n c e, t h e t o t a l n u m b e r o f w a y s = 780 \end{array}$

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Exemplar Solution Class 11th

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!]

23 Views|Posted a year ago

Asked by Payal Gupta

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Answered by

Payal Gupta | Contributor-Level 10

a year ago

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

$\begin{array}{l} T h e a l p h a b e t i c a l o r d e r o f R A C H I T i s A, C, H, I, R a n d T \\ N u m b e r o f w o r d s b e g i n n i n g w i t h A = 5! \\ N u m b e r o f w o r d s b e g i n n i n g w i t h C = 5! \\ N u m b e r o f w o r d s b e g i n n i n g w i t h H = 5! \\ N u m b e r o f w o r d s b e g i n n i n g w i t h I = 5! \\ a n d N u m b e r o f w o r d s b e g i n n i n g w i t h R i . e . R A C H I T = 1 \\ ∴ T h e r a n k o f t h e w o r d' R A C H I T' i n t h e d i c t i o n a r y \\ = 5! + 5! + 5! + 5! + 1 = 4 \times 5! + 1 \\ = 4 \times 5.4.3.2.1 + 1 = 4 \times 120 + 1 = 480 + 1 = 481 \end{array}$

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Exemplar Solution Class 11th

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.]

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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 short answer type question as classified in NCERT Exemplar

$\begin{array}{l} W e h a v e 2 w o m e n a n d 3 m e n \\ F i r s t w o m e n c h o o s e t h e c h a i r s a m o n g s t t h e c h a i r s 1 t o 4 i . e . \\ T o t a l n u m b e r o f c h a i r s = 4 \\ S o, t h e n u m b e r o f a r r a n g e m e n t s = {}_{4}{P_{2}} w a y s \\ N o w 3 m e n c h o o s e f r o m t h e r e m a i n i n g 6 c h a i r s \\ S o, t h e n u m b e r o f a r r a n g e m e n t s = {}_{6}{P_{3}} w a y s \\ ∴ T o t a l n u m b e r o f a r r a n g e m e n t s = {}_{4}{P_{2}} \times {}_{6}{P_{3}} \\ = \frac{4!}{(4 - 2)!} \times \frac{6!}{(6 - 3)!} = \frac{4!}{2!} \times \frac{6!}{3!} \\ = \frac{4.3.2!}{2!} \times \frac{6.5.4.3!}{3!} = 12 \times 120 = 1440 \\ H e n c e, t h e t o t a l n u m b e r o f p o s s i b l e a r r a n g e m e n t s a r e 1440 . \end{array}$

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Exemplar Solution Class 11th

Choose the correct option for $x$ :

(a) $x \in (- \infty, - 2)$  

(b) $x \in (- \infty, - 2]$  

(c) $x \in (- 2, \infty]$  

(d) $x \in [- 2, \infty)$

4 Views|Posted a year ago

Asked by Payal Gupta

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Answered by

Payal Gupta | Contributor-Level 10

a year ago

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l} T h e g i v e n g r a p h r e p r e s e n t s a l l r e a l v a l u e s o f x l e s s t h a n a n d e q u a l t o - 2 . \\ S o, x \in (- \infty, - 2] \\ H e r e, t h e c o r r e c t o p t i o n i s (b) . \end{array}$

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Exemplar Solution Class 11th

Choose the correct option for $x$ :

(a) $x \in (- \infty, \frac{7}{2})$  

(b) $x \in (- \infty, \frac{7}{2}]$  

(c) $x \in [\frac{7}{2}, \infty)$  

(d) $x \in (\frac{7}{2}, \infty)$

3 Views|Posted a year ago

Asked by Payal Gupta

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Answered by

Payal Gupta | Contributor-Level 10

a year ago

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l} T h e g i v e n g r a p h r e p r e s e n t s a l l t h e v a l u e s o f x l e s s t h a n \frac{7}{2} o n t h e r e a l n u m b e r l i n e . \\ S o, x \in (- \infty, \frac{7}{2}) \\ H e r e, t h e c o r r e c t o p t i o n i s (a) . \end{array}$

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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.

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

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

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?

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.]

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. Find the number of different ways of doing questions.

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!]