Question

How many distinct teams of 3,4&5 can be formed from a group of 12?

Answer 1

A team of 3 would be picked like this: for the 1st team member we have a choice of 12, for the 2nd a choice of 11 and for the 3rd a choice of 10. So that gives us 12*11*10 possible choices taking into account the order. If the order doesn't matter then we need to divide this by 6, because there are 6 different ways of arranging 3 team members. Therefore there are 220 ways of picking a team of 3. To pick a team of 4 it's 12*11*10*9 divided by 24=495; and for a team of 5 it's 12*11*10*9*8 divided by 120=792.

If the team of 4 is to be picked from the 9 remaining people after team 3 has been picked, then for this team we only have a choice of 9, so the calculation is 9*8*7*6/24=126; and to pick a team of 5 from 5 remaining people, well there's only one way! So that' s 220*126*1=27720, when the order of selecting teams is 3, 4 and 5.

If the order of selection is 4, 3 and 5, it's 495 then 8*7*6/6=56, which makes 495*56*1=27720 also.

If the order is 5, 4, 3, it's 792 then 7*6*5*4/24=35, making 792*35*1=27720. So the order makes no difference.

 

Answer 2

When 3 people are chosen, e.g. A, B and C, from 12 members, the order of 3, e.g. ABC, ACB, BAC, ... , doesn't matter.   So, we can use the formula for combinations, the combination of "n" objects taken "r" at a time: C(n,r)=n!/r!(n-r)!.

(i). If 3 are chosen from12: C(12,3)=12!/3!(12-3)!=220    We have 220 possible combinations.

(ii). If 4 are chosen from the rest of 9: C(9,4)=9!/4!(9-4)=126  We have 126 possible combinations.

(iii). If 5 are chosen from the rest of 5: C(5,5)=5!/5!(5-5)=1   We have only one choice left.

Thus, we have: C(12,3)xC(9,4)xC(5,5)=220x126x1=27720*

The answer: 27720 possible combinations.

Even if we change the order of selection, 3, 4 and 5, the number of possible combinations won't be changed, e.g. C(12,4)xC(8,5)xC(3,3)=495x56x1=27720 (proving this is skipped).