Question

A school committee consists of 2 teachers and 4 students. What is the number of different committees that can be formed from 5 teachers and 10 students?
(a) 10
(b) 15
(c) 2100
(d) 8

Answer 1

Hint: This is a combination problem of choosing 2 teachers and 4 students from 5 teachers and 10 students. Use the combination formula to find the number of ways of choosing 2 teachers from 5 teachers and 4 children from 10 children and multiply them to find the answer.

Complete step-by-step answer:
It is given that the school committee consists of 2 teachers and 4 students. The number of available teachers is 5 and the number of available students is 10.
Therefore, we need to choose 2 teachers from 5 teachers and 4 students from 10 students.
The number of ways of choosing two teachers from five teachers is ${}^5{C_2}$.
We know the formula for ${}^n{C_r}$ as follows:
${}^n{C_r} = \dfrac{{n!}}{{r!(n - r)!}}................(1)$
Calculating the value of ${}^5{C_2}$ from equation (1), we have:
$\Rightarrow$ ${}^5{C_2} = \dfrac{{5!}}{{2!(5 - 2)!}}$
Simplifying, we get:
$\Rightarrow$ ${}^5{C_2} = \dfrac{{5!}}{{2!3!}}$
$\Rightarrow$ ${}^5{C_2} = 10...........(2)$
The number of ways of choosing four students out of ten students is ${}^{10}{C_4}$.
The value of ${}^{10}{C_4}$ from equation (1) is given as follows:
$\Rightarrow$ ${}^{10}{C_4} = \dfrac{{10!}}{{4!(10 - 4)!}}$
Simplifying, we obtain as follows:
$\Rightarrow$ ${}^{10}{C_4} = \dfrac{{10!}}{{4!6!}}$
$\Rightarrow$ ${}^{10}{C_4} = 210.............(3)$
Therefore, the number of ways of choosing 2 teachers and 4 students from 5 teachers and 10 students is $\Rightarrow$ ${}^5{C_2} \times {}^{10}{C_4}$.
From equation (2) and equation (3), we have:
$\Rightarrow$ ${}^5{C_2} \times {}^{10}{C_4} = 10 \times 210$
$\Rightarrow$ ${}^5{C_2} \times {}^{10}{C_4} = 2100$
Hence, the correct answer is option (c).

Note: You can also consider it as a permutation of 5 teachers into two similar groups having 2 and 3 teachers respectively and similarly with the 10 children into two similar groups of 4 and 6 children respectively and solve the problem.