Question

(i) If 20C12 = 20Cr
then one of the possible value of r = _________.
(a) 9 (b) 8 (c) 11 (d) 10 ...(1 Mark)

(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 Mark)

Answer 1

Step-by-step explanation:

Given:

(i) If 20C12 = 20Cr

To find: one of the possible value of r = _________.

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

Solution:

Tip:Formula for combination

$\bold{^nC_r=\frac{n!}{r!(n-r)!}}$

Apply formula of combination

$\frac{20!}{12!(20 - 12)!} = \frac{20!}{r!(20 - r)!}$

Cancel common term and cross multiply

12!(20-12)!=r!(20-r)!

now think for a value of r for which the equation balances.

if we put r=8

then

12!8!=8!12!

Thus,both are equal.

So,

Option (b) is correct.

(ii)Given: A bag contains 7 red balls and 5 white balls.

To find: Determine the number of ways in

which 3 red balls and 2 white balls can be selected.

Solution:

Tip:Formula for permutation

$\bold{^nP_r=\frac{n!}{(n-r)!}}$

3 red balls are selected from 7 red balls,thus number of ways 7P3

2 white balls are selected from 5 white balls,thus number of ways 5P2

Thus, total ways

$=\frac{7!}{(7 - 3)!} + \frac{5!}{(5 - 2)!}$

$=\frac{7!}{4!} + \frac{5!}{3!}$

$=\frac{7 \times 6 \times 5 \times 4!}{4!} + \frac{5 \times 4 \times 3!}{3!}$

= 7×6×5+5×4

=210+20

=230 ways

Final answer:

1) r=8

Option b is correct.

2)No. of possible ways of choosing red and white balls as per given conditions is 230.

Hope it helps you.

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Answer 2

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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