Question

in how many ways 2 boys and 1 girl can be selected from a class of 10 boys and 20 girls​

Answer 1

The number of different ways in which 2 boys and 1 girl can be selected from a class of 10 boys and 20 girls​ = 900.

Given:

A class of 10 boys and 20 girls.

To Find:

The number of ways in which 2 boys and 1 girl can be selected from a class of 10 boys and 20 girls​.

Solution:

To solve this problem we will use the concept of combinations.

The number of ways in which 'r' items can be chosen from a total of 'n' items is represented by $\limits^n {C}_{r}$ = $\frac{n!}{(n-r)!&\ r!}$

It is given that

The number of boys in the class = 10

The number of ways in which 2 boys can be selected from a total of 10 boys is denoted by $\limits^{10} {C}_{2}$ = $\frac{10!}{(10-2)!&\ 2!}$ = $\frac{10!}{8!&\ 2!}$ = 45

The number of girls in the class = 20

The number of ways in which 1 girl can be selected from a total of 20 girls is denoted by $\limits^{20} {C}_{1}$ = $\frac{20!}{(20-1)!&\ 1!}$ = 20.

Hence, the number of ways in which 2 boys and 1 girl can be selected from a class of 10 boys and 20 girls​ =  $\limits^{10} {C}_{2}$ x $\limits^{20} {C}_{1}$ = 45 x 20 = 900.

∴ The number of different ways in which 2 boys and 1 girl can be selected from a class of 10 boys and 20 girls​ = 900.

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Answer 2

Answer:

In $900$ ways.

Step-by-step explanation:

Formula for number of combination,

$n_C_r=\frac{n!}{r!(n-r)!}$

where $n=$ total no. things in a set and $r=$ no. of selecting things from the given set.

Total no. of boys in a class = $10$

Total no. of girls in a class = $20$

According to the question,

From class of $10$ boys, no. of boys selected = $2$

So, the no. of ways of selection of $2$ boys = ${10}_C_2$

= $\frac{10!}{2!(10-2)!}$

= $\frac{10!}{2!8!}$

= $\frac{10\times 9\times 8!}{2!\ 8!}$

= $45$ ways

Also, From class of $20$ girls, no. of girls selected = $1$

So, the no. of ways of selection of $1$ girl = ${20}_C_1$

= $\frac{20!}{1!(20-1)!}$

= $\frac{20!}{1!19!}$

= $\frac{20\times 19!}{1!\ 19!}$

= $20$ ways

The total number of ways of selecting $2$ boys and $1$ girl = $45\times 20$ ways

                                                                                            = $900$ ways

Therefore, from a class of $10$ boys and $20$ girls​, in $900$ ways $2$ boys and $1$ girl can be selected.

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