Question

(i) If 20C12 = 20Cr

then one of the possible value of r = _________.

(a) 9 (b) 8 (c) 11 (d) 10 (1)

(ii) A bag contains 7 red balls and 5 white balls. Determine the number of ways in

which 3 red balls and 2 white balls can be selected. (3)​

Answer 1

SOLUTION

TO DETERMINE

$\sf (i) \: If \: \: {}^{20} C_{12} = {}^{20} C_{r}$

then one of the possible value of r = ____

(ii) A bag contains 7 red balls and 5 white balls. Determine the number of ways in which 3 red balls and 2 white balls can be selected

EVALUATION

(i) Here it is given that

$\sf {}^{20} C_{12} = {}^{20} C_{r}$

We now implement the formula

$\boxed{ \: \sf {}^{n} C_{r} = {}^{n} C_{n - r} \: \: }$

$\sf {}^{20} C_{12} = {}^{20} C_{r}$

$\implies \sf {}^{20} C_{20 - 12} = {}^{20} C_{r}$

$\implies \sf {}^{20} C_{8} = {}^{20} C_{r}$

Comparing both sides we get r = 8

Hence the required value of r = 8

(ii) Total number of red balls = 7

The number of ways in which 3 red balls can be selected

$\sf = {}^{7} C_{3}$

$\sf = \dfrac{7!}{3!(7 - 3)!}$

$\sf = \dfrac{7!}{3!4!}$

$\sf = \dfrac{7 \times 6 \times 5}{3 \times 2}$

$\sf = 35$

Total number of white balls = 5

The number of ways in which 2 white balls can be selected

$\sf = {}^{5} C_{2}$

$\sf = \dfrac{5!}{2!(5 - 2)!}$

$\sf = \dfrac{5!}{2!3!}$

$\sf = \dfrac{5 \times 4 }{2}$

$\sf = 10$

Hence the required number of ways

= 35 × 10

= 350

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Learn more from Brainly :-

$1.$ If 7. 5pr = 7pr – 1, then find r.(question based on permutation and combination chapter)

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2. If nPr = 3024 and nCr = 126 then find n and r.

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Answer 2

Step-by-step explanation:

SOLUTION

TO DETERMINE

$\sf (i) \: If \: \: {}^{20} C_{12} = {}^{20} C_{r}$

then one of the possible value of r = ____

(ii) A bag contains 7 red balls and 5 white balls. Determine the number of ways in which 3 red balls and 2 white balls can be selected

EVALUATION

(i) Here it is given that

$\sf {}^{20} C_{12} = {}^{20} C_{r}$

We now implement the formula

$\boxed{ \: \sf {}^{n} C_{r} = {}^{n} C_{n - r} \: \: }$

$\sf {}^{20} C_{12} = {}^{20} C_{r}$

$\implies \sf {}^{20} C_{20 - 12} = {}^{20} C_{r}$

$\implies \sf {}^{20} C_{8} = {}^{20} C_{r}$

Comparing both sides we get r = 8

Hence the required value of r = 8

(ii) Total number of red balls = 7

The number of ways in which 3 red balls can be selected

$\sf = {}^{7} C_{3}$

$\sf = \dfrac{7!}{3!(7 - 3)!}$

$\sf = \dfrac{7!}{3!4!}$

$\sf = \dfrac{7 \times 6 \times 5}{3 \times 2}$

$\sf = 35$

Total number of white balls = 5

The number of ways in which 2 white balls can be selected

$\sf = {}^{5} C_{2}$

$\sf = \dfrac{5!}{2!(5 - 2)!}$

$\sf = \dfrac{5!}{2!3!}$

$\sf = \dfrac{5 \times 4 }{2}$

$\sf = 10$

Hence the required number of ways

= 35 × 10

= 350

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