There are 16C6 = 16*15*14*13*12*11/(6*5*4*3*2*1) possible committees.
That is, 8,008 of them.
72
There are 4845 ways to choose 4 people out of 20 20 choose 4 = 20! / (4!16!)
-5
The number of distinct, different 4-person committees that can be formedfrom a group of 32 people is(32!/28!) / (4!) = (32 x 31 x 30 x 29) / (4 x 3 x 2 x 1) = 35,960(but obviously, no more than four at a time.)
Possibilities are (9 x 8)/2 times (49 x 48 x 47 x 46)/24 = 366,121,728/48 =7,627,536 different committees.
120
There are: 10C7 = 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.
To determine the number of different two-person committees that can be formed from a group of six people, we use the combination formula ( C(n, k) = \frac{n!}{k!(n-k)!} ), where ( n ) is the total number of people and ( k ) is the number of people to choose. Here, ( n = 6 ) and ( k = 2 ). Thus, the number of combinations is ( C(6, 2) = \frac{6!}{2!(6-2)!} = \frac{6 \times 5}{2 \times 1} = 15 ). Therefore, 15 different two-person committees can be formed.
72
There are 4845 ways to choose 4 people out of 20 20 choose 4 = 20! / (4!16!)
-5
There are 2300 possible combinations.
it has 20 people in the standing committee not how many people how many committees
To determine how many 5-person committees can be formed from nine people, you can use the combination formula ( C(n, k) = \frac{n!}{k!(n-k)!} ), where ( n ) is the total number of people, and ( k ) is the number of people to choose. In this case, ( n = 9 ) and ( k = 5 ). Thus, the number of ways to select the committee is ( C(9, 5) = \frac{9!}{5!(9-5)!} = \frac{9!}{5!4!} = 126 ). Therefore, there are 126 different 5-person committees possible.
(9 x 8 x 7 x 6)/(4 x 3 x 2 x 1) = 126committees.
Eight. If you can't multiply by fives, then your either an idiot, or too young to be using the Internet.