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
3 students can be chosen from a class of 30 in (30 x 29 x 28) = 24,360 ways.But each group of the same 3 students will be chosen in six different ways.The number of different groups of 3 is 24,360/6 = 4,060 .
30C8 = 5,852,925
6,375,600
20 x 19 x 18/3 x 2 = 1,140 groups
There are 11880 ways.
There are 10 different sets of teachers which can be combined with 4 different sets of students, so 40 possible committees.
There are 247 groups comprising 2 or more students.
they can be 2 groups of 16, 4 groups of 8, 8 groups of 4, or 16 groups of 2
2.33333333
Any 5 from 7 is (7 x 6)/2 ie 21.
two
C105 = 10 ! / (5!x(10-5)!) = 10! /5!2 = 252