Question

In how many ways 2 boys and 2 girls can be selected from a class of 10 boys and 5 girls?

Answer 1

BOYS:

6C3

6!/3!*(6!-3!) = 6!/3!*3!

(6*5*4*3*2*1)/(3*2*1*3*2*1) = 20 ways

GIRLS:

5C2

5!/2!*(5-2)! = 5!/2!*3!

(5*4*3*2*1) / (2*1*3*2*1) = 10 ways

The combined effort would therefore be 10*20 ways = 200 ways.

Answer 2

There are 450 different ways of choosing 2 boys and 2 girls from a class of 10 boys and 5 girls.

Given:

In a class, there are 10 boys and 5 girls.

To Find:

The number of ways in which 2 boys and 2 girls can be selected from a class of 10 boys and 5 girls.

Solution:

In order to solve this problem, we will use the concept of combinations.

The number of ways in which 'r' items can be chosen out of a total of 'n' items is given by $\limits^nC_ {r}$ = $\frac{n!}{(n-r)!\,r!}$

Total number of boys in the class, n = 10

The number of boys to be chosen, r = 2.

So, the number of ways in which 2 boys can be selected from a class of 10 boys = $\limits^{10}C_ {2}$ = $\frac{10!}{(10-2)!\,2!}$ = 45.

∴ There are 45 different ways of choosing 2 boys out of a total of 10 boys.

Total number of girls in the class, n = 5

The number of girls to be chosen, r = 2.

So, the number of ways in which 2 boys can be selected from a class of 5 girls = $\limits^{5}C_ {2}$ = $\frac{5!}{(5-2)!\,2!}$ = 10.

∴ There are 10 different ways of choosing 2 girls out of a total of 5 girls.

Now, the total number of ways in which 2 boys and 2 girls can be selected from a class of 10 boys and 5 girls =  $\limits^{10}C_ {2}$ x  $\limits^{5}C_ {2}$ = 45 x 10 = 450.

∴ There are 450 different ways of choosing 2 boys and 2 girls from a class of 10 boys and 5 girls.

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