In how many ways can $3$ boys and $2$ girls be selected from a group of $6$ boys and $4$ girls?

Question

There is a group of $6$ boys and $4$ girls.

In how many ways can $3$ boys and $2$ girls be selected from a group of $6$ boys and $4$ girls?

A. $60$

B. $80$

C. $120$

D. $160$

I tried with $^nC_k$ but I couldn't do it. This is what I tried: $6!/3!*3!=20$

Answer 1

$3$ boys can be chosen from $6$ boys in $^6C_3$ ways $=\binom{6}{3}$ways

$2$ girls can be chosen from $4$ girls in $^4C_2$ ways $=\binom{4}{2}$ways

Now, the number of combinations $$=\ ^6C_3\cdot^4C_2$$ $$=\binom{6}{3}\cdot\binom{4}{2}$$ $$=\left(\frac{6!}{3!(6-3)!}\right)\cdot\left(\frac{4!}{2!(4-2)!}\right)$$ $$=120$$ Total number of ways $=120$

Answer 2

Essentially we are asking how many groups of $3$ boys from a pool of $6$ boys are there - answer $^6C_3=\frac{6!}{3!3!}=20$

Similarly, there $^4C_2=\frac{4!}{2!2!}=6$ possible groups of girls. Since these are independent, the total number is $20$ x $6$ $= 120$.

Answer = $120$.