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).