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
To divide 32 students into groups with equal numbers of students, the group sizes must be divisors of 32. The divisors of 32 are 1, 2, 4, 8, 16, and 32. Therefore, the students can be grouped into 1 group of 32, 2 groups of 16, 4 groups of 8, 8 groups of 4, 16 groups of 2, or 32 groups of 1.
To find the number of different groups of 3 students that can be selected from 10 students, we use the combination formula: ( C(n, r) = \frac{n!}{r!(n-r)!} ). Here, ( n = 10 ) and ( r = 3 ), so the calculation is ( C(10, 3) = \frac{10!}{3!(10-3)!} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 120 ). Therefore, there are 120 different groups of 3 students that can be formed.
30C8 = 5,852,925
6,375,600
20 x 19 x 18/3 x 2 = 1,140 groups
There are 11880 ways.
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.
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