if order does not matter then, (23x22x21x20x19)/(5x4x3x2x1) = 33,649
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20 x 19 x 18/3 x 2 = 1,140 groups
two
There are only two possibilities... 10 groups of 2 or 5 groups of 4. Unless - you can have varying sized groups - which you didn't specify.
The first member chosen can be any one of 1,514 students.The second member chosen can be any one of the remaining 1,513 students.The third member chosen can be any one of the remaining 1,512 students.So there are (1,514 x 1,513 x 1,512) ways to choose three students.But for every group of three, there are (3 x 2 x 1) = 6 different orders in which the same 3 can be chosen.So the number of `distinct, unique committees of 3 students is(1514 x 1513 x 1512) / 6 = 577,251,864
we can make 1 group