There are 4845 ways to choose 4 people out of 20
20 choose 4 = 20! / (4!16!)
There are 16C6 = 16*15*14*13*12*11/(6*5*4*3*2*1) possible committees. That is, 8,008 of them.
72
-5
I think there are 88 different combinations of coins that can make up 66 cents.
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.)
120
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.
Any combination of 5 students leaves one student out. Since there are 5 possible students to leave out, the number of combinations of all but one student is 5.
There are 16C6 = 16*15*14*13*12*11/(6*5*4*3*2*1) possible committees. That is, 8,008 of them.
Since the order doesn't matter, this is a combination problem.25C3 = 25*24*23/3*2*1 = 25*4*23 = 2300 committees.
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.
(9 x 8 x 7 x 6)/(4 x 3 x 2 x 1) = 126committees.
72
9 combinations - the key person and one of the remaining nine.
-5
There is no set number. It could be none at all.
The person who chooses the chairs for each of the standing committees in the Assembly is typically the Speaker of the Assembly. The Speaker has the authority to appoint committee chairs, often considering factors such as party representation, seniority, and expertise in relevant areas. This role is crucial for ensuring that committees function effectively and align with the legislative agenda.