how many ways 3 pairs can be selected out of 6 students?

Question

Out of "6 students" how many ways can 3 pairs be selected for assigning homework ?

Answer 1

Answer:

You select the first pair by ${6\choose2}$ number of ways, the next pair by ${4\choose2}$ and the third pair by ${2\choose2}$. So the total number of ways you can select three pairs for assigning homework is $\left({6\choose2}{4\choose2}{2\choose2}\right)/3!$ = 15 ways. You divide by 3! to eliminate order.

Answer 2

Selecting a pair means choosing 2 students out of the collection.

Let us choose first pair: We have 6 students and so this can be done in $6\choose 2$ ways.

Now time to choose the second pair: We are left now with 4 students and so it can be done in $4\choose 2$ ways.

Thus last pair can be chosen in $2\choose 2$ ways

So it seems that required number of ways of choosing is $6\choose 2$ $\times$ $4\choose 2$ $\times$ $2\choose 2$

But wait,there are repeatations: So to avoid choosing same "3 pairs" more than once,we have to divide it by total number of permutations of 3 pairs which is $3!=6$.

Hence answer is $\frac{1}{6}$$6\choose 2$ $\times$ $4\choose 2$ $\times$ $2\choose 2$