if order does not matter then, (23x22x21x20x19)/(5x4x3x2x1) = 33,649
20 x 19 x 18/3 x 2 = 1,140 groups
The number of different groups of students that can be formed from 16 students depends on the size of the groups being formed. If you are looking for all possible combinations of groups of any size (from 1 to 16), you can use the formula for combinations. The total number of combinations would be (2^{16} - 1) (subtracting 1 to exclude the empty group), which equals 65,535 different groups. If you specify a particular group size, the calculation would be different.
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.
To find the number of different groups of 4 that can be made from 17 students, you can use the combination formula, which is given by ( C(n, r) = \frac{n!}{r!(n-r)!} ). In this case, ( n = 17 ) and ( r = 4 ). Therefore, the calculation is ( C(17, 4) = \frac{17!}{4!(17-4)!} = \frac{17!}{4! \times 13!} = \frac{17 \times 16 \times 15 \times 14}{4 \times 3 \times 2 \times 1} = 2380 ). Thus, there are 2,380 different groups of 4 that can be formed from 17 students.
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.
Well, honey, there are 30 students in the class, and you want to choose a group of 3. So, you're looking at a classic combination situation. The formula for combinations is nCr = n! / r!(n-r)!, so in this case, it's 30C3 = 30! / 3!(30-3)! = 4060 ways to choose those 3 lucky students. It's like picking the winning lottery numbers, but with fewer tears and more math.
2.33333333
Any 5 from 7 is (7 x 6)/2 ie 21.
two
C105 = 10 ! / (5!x(10-5)!) = 10! /5!2 = 252
Two