Question

In how many ways can a team of 3 boys and 3 girls be selected from 5 boys and 4 girls?

Answer 1

Hint: First find the number of ways 3 girls will be selected from 4 girls. Then, find the number of ways to select 3 boys out of 5 boys. After that for a team of 3 boys and 3 girls multiply the number of ways boys have been selected with the number of ways girls must be selected.

Formula used:
The number of selections of r objects from the given n objects is given by $^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$.

Complete step-by-step answer:
Given:-
There are 4 girls out of which 3 girls have to be selected. So,
No of ways =$^n{C_r}$
Put, n= 4 and r= 3,
$^4{C_3} = \dfrac{{4!}}{{3!\left( {4 - 3} \right)!}}$
Subtract 3 from 4 in the bracket we get,
$^4{C_3} = \dfrac{{4!}}{{3!1!}}$
Now, expand the factorial,
$^4{C_3} = \dfrac{{4 \times 3 \times 2 \times 1}}{{3 \times 2 \times 1 \times 1}}$
Cancel out the common factors to get the number of ways 3 girls can be selected,
$^4{C_3} = 4$ …..(1)
There are 5 boys out of which 3 boys have to be selected. So,
No of ways =$^n{C_r}$
Put, n= 5 and r= 3,
$^5{C_3} = \dfrac{{5!}}{{3!\left( {5 - 3} \right)!}}$
Subtract 3 from 5 in the bracket we get,
$^5{C_3} = \dfrac{{5!}}{{3!2!}}$
Now, expand the factorial,
$^5{C_3} = \dfrac{{5 \times 4 \times 3 \times 2 \times 1}}{{3 \times 2 \times 1 \times 2 \times 1}}$
Cancel out the common factors to get the number of ways 3 boys can be selected,
$^5{C_3} = 10$ …..(2)
So, the total number of ways to select 3 boys and 3 girls are the product of the number of ways 3 boys must be selected with the number of ways 3 girls must be selected. Then,
$Total\,ways{ = ^4}{C_3}{ \times ^5}{C_3}$
Substitute the values from the equation (1) and (2),
$Total\,ways = 4 \times 10$
Multiply the terms on the right side to get the final result,
$Total\,ways = 40$

Hence, the total number of ways to select 3 boys and 3 girls is 40.

Note: The students are likely to make mistakes when finding the number of ways of selection of the team.
$Total\,ways{ = ^4}{C_3}{ \times ^5}{C_3}$
A combination is a mathematical technique that determines the number of possible arrangements in a collection of items where the order of the selection does not matter. In combinations, you can select the items in any order.
The formula for determining the number of possible arrangements by selecting only a few objects from a set with no repetition is expressed in the following way:
$C\left( {n,r} \right) = \left( \begin{gathered}
  n \\
  k \\
\end{gathered} \right) = \dfrac{{n!}}{{k!\left( {n - k} \right)!}}$