# Thread: 8 boys and 10 girls in the class

1. ## 8 boys and 10 girls in the class

There are 8 boys and 10 girls in the class. Two pairs (2 boys and 2 girls) must be selected for certain purposes. What is the number of different selections?

My solution is: C(2,8) * C(2,10)

My friend says: (C(2,8) * C(2,10)) / C(4,18)

which one is correct and WHY? Thank you

2. You are correct. You can choose 2 out of 8 for boys and 2 out of 10 for girls.

3. Originally Posted by Narek
There are 8 boys and 10 girls in the class. Two pairs (2 boys and 2 girls) must be selected for certain purposes. What is the number of different selections?

My solution is: C(2,8) * C(2,10)

My friend says: (C(2,8) * C(2,10)) / C(4,18)

which one is correct and WHY? Thank you
The statement of the problem is, in my opinion, ambiguous. Does it matter who is paired with whom (as would be the case for dance partners, for example), or do we just need 2 boys and 2 girls without any pairing?

Personally, I think the statement that we need two pairs indicates that the pairings are significant. In that case, you need to multiply your answer by 2. Do you see why?

4. As usual I disagree with all the responses.
I do agree that there should be two distinct pairs that are disjoint.
The first pair can be selected in $\displaystyle 8\times 10$ ways.
Then the second pair can be selected in $\displaystyle 7\times 9$ ways.
Now we multiply those two. BUT we divide by two to avoid duplication.

5. If I understand Plato correctly, then, we agree on the final answer:

C(8,2) * C(10,2) * 2 = 2520

8 * 10 * 7 * 9 / 2 = 2520