#1 (permalink)

4 boys and 7 girls. In how many ways can a team of 5 be selected if:

(a) The team was to have no boys.
(b) The team had to have a member of each sex
(c) The team had to have 2 boys and 3 girls
(d) Atleast 3 boys had to be on a team


Help me understand the reasoning behind the answers to each one. Inevitably I'm going to get them tomorrow when the teacher calls out the answer, but he'll pretty much never explain them :3

#2 (permalink)

The reasoning is that it is combinations... like for part a. There can be no boys so ignore the boys so that leaves 7 girls. Team of 5 and there are 7 people so you use that formula but use the formula for unique number of teams since like team abcde is the same as abced or decab or any combination of that (where letters stand for people). The other formula would count all of those duplicates and would give a higher number. I haven't done this in a while so I can't say exactly the answers or anything without relearning but it's easy to plug into a formula.
The formulas I'm referring to are like npr and ncr or something... The one that gives smaller numbers is the unique combination one while the other one counts all combinations including duplicates that are in different orders.

b. Apply the formula for all unique combinations of 11 people in teams of 5. Then from that answer subtract away the cases where you have all 5 girls (take 7 people in teams of 5 and see how many unique combinations).

c. Think of it as a two teams: a team of 2 (boys only) and a team of 3 (girls only). So you can do the formula for each one then multiply the answers together I believe. Like for boys a, b, c, d. The unique combinations are ab, ac, ad, bc, bd, cd. 6 of them then for girls you do the same (again with the formula or if you feel like it, by hand). Then put them together for each like ab with 123 then ac with 123 then ad with 123 (so it's the same as multiplying the number of unique combinations together).

d. Take all possible unique combinations of 11 people on teams of 5. Then subtract away all combinations where the number of boys on a team is less than 3. How can you get that second number? Maybe there is an easier way that I'm not thinking of but you could add up all the ones where there are two boys only (answer in part c) and one boy only (do same as part c except with one boy and four girls) and no boys (calculated in part a).

I haven't done this stuff for years so I'm just saying it from logic which is what I think you wanted. If any part isn't clear or is wrong let me know, thanks.

#3 (permalink)

Actually your reasoning sounds very...sound (heh), and goes with the answers my teacher gave us. The only one that was wrong was part d. Instead, we had to get combinations with 3 boys and 2 girls, then the ones with 4 boys and 1 girl. Then we add them together.

Thanks for your insight into these questions. I think I get it a LITTLE better now... This topic is a little tricky though, and I lack logical skills but I think I see a little better now what we're trying to achieve and how we got there.

#4 (permalink)

Oh yeah I read d wrong (mine would have been at most 3 boys).

Np. It's just making teams/combinations and in this case it goes by unique combinations since it doesn't matter what order they are in, but say the question asked all the combinations 5 people could have come in a race - then the order would matter since 1st place, 2nd place, etc makes a difference.