72
15
There are 16C6 = 16*15*14*13*12*11/(6*5*4*3*2*1) possible committees. That is, 8,008 of them.
The answer is 15 * 14 * 13 = 15 P 3 = 780 IF you assume that no one person can hold two offices at once and that all in the group are qualified for any office.
5 for 2, 3 for 3, 2 for 4.
72
84 = 9 choose 3 = 9! / 3! / (9-3)!
-5
Since the order doesn't matter, this is a combination problem.25C3 = 25*24*23/3*2*1 = 25*4*23 = 2300 committees.
That depends on what the joint committee is. Joint committees refers to a wide group of committees that share a common characteristic, that they have membership from both the house and the senate. Many joint committees are standing committees, which are permanent. At the same time, many joint committees arent permanent. So pretty much: Joint committees can be permanent, but dont have to be.
There are: 10C7 = 120
The answer is 7C5 = 21.
120
They can't be split evenly into groups of six. Sixteen people can split into two groups of six, and there will be four people left over.
There is no set number. It could be none at all.
There are 2300 possible combinations.
(9 x 8 x 7 x 6)/(4 x 3 x 2 x 1) = 126committees.