The answer is 7C5 = 21.
There are: 10C7 = 120
-5
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.
To determine how many 3-member committees can be formed from a group of 18 students, you can use the combination formula: (C(n, r) = \frac{n!}{r!(n-r)!}), where (n) is the total number of students and (r) is the number of members in the committee. In this case, (n = 18) and (r = 3). Thus, the calculation is (C(18, 3) = \frac{18!}{3!(18-3)!} = \frac{18 \times 17 \times 16}{3 \times 2 \times 1} = 816). Therefore, you can form 816 different 3-member committees from the group of 18 students.
(9 x 8 x 7 x 6)/(4 x 3 x 2 x 1) = 126committees.
There are 2300 possible combinations.
-5
5 for 2, 3 for 3, 2 for 4.
72
There are: 10C7 = 120
-5
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.
no it means a group of people that may have different specialties within the group
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.)
To determine how many 3-member committees can be formed from a group of 18 students, you can use the combination formula: (C(n, r) = \frac{n!}{r!(n-r)!}), where (n) is the total number of students and (r) is the number of members in the committee. In this case, (n = 18) and (r = 3). Thus, the calculation is (C(18, 3) = \frac{18!}{3!(18-3)!} = \frac{18 \times 17 \times 16}{3 \times 2 \times 1} = 816). Therefore, you can form 816 different 3-member committees from the group of 18 students.
Six
They are called `lobbyists`.