Question 1205064: There are 12 members on a board of directors. If they must form a subcommittee of members 3, how many different subcommittees are possible?
Found 2 solutions by math_helper, math_tutor2020:
Answer by math_helper(2461)   (Show Source):
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Assuming the 3 subcommittee members don't have titles (i.e. they serve with equal standing and none are distinguishable from each other with respect to the subcommittee):
C(12,3) = 12!/((12-3)!*3!) = 12!/(9!*3!) = 220
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Aside...
If there was a "head of subcommittee" and "spokesperson of subcommittee" then we would not need to divide by the 3! above (similar to "order matters") and you'd get:
P(12,3) = 12!/(12-3)! = 12!/9! = 1320
Aside #2...
If, say, exactly one member was 'spokesperson' and the other two were otherwise equal (without title), you'd have:
12 * C(11,2) = 12*(11!/(9!*2!)) = 660
You can see from the pattern, the more distinctions that are made, the greater the number of subcommittees that can be formed.
Answer by math_tutor2020(3816)  (Show Source):
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Answer: 220
Explanation
There are 12*11*10 = 1320 different permutations of 3 members if order mattered. On a committee without seat labels (eg: "president", "VP", etc) the order doesn't matter. We divide by 3! = 3*2*1 = 6 since this is the number of ways to rearrange a trio of people.
1320/6 = 220 is the number of combinations of 3 people where order doesn't matter. A group like {A,B,C} is the same as {B, C, A}.
Another way to reach this answer is to use the nCr combination formula with n = 12 and r = 3.
Or you can look at Pascal's triangle. Look at the row that starts with 1,12,... and count 3 spots to the right to land on 220
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