In a given city there are 28 reporters of which 16 are men. If the committee of 4 reporters is to be formed
A. How many different committees can be formed?
B. How many different committees contain 3 men?
C. How many different commitrees contain at least 2 men?
D. How many different committees contain at most 3 men?
16 men and 12 women
A - you want 4 from 28 , so C(28,4)
B - 3 men, must be 1 woman, so C(16,3)*C(12,1)
C - at least 2 men ---> 2men-2women , 3men1woman, 4men , add up the 3 cases
D - your turn
Thank you :)
To solve these counting problems, we can use combinations and permutations.
A combination is the selection of items from a larger set without considering the arrangement or order of the items. The formula for combinations is given by:
C(n, r) = n! / (r!(n-r)!)
where n is the total number of items and r is the number of items chosen.
A permutation, on the other hand, takes into account the arrangement or order of the items. The formula for permutations is given by:
P(n, r) = n! / (n-r)!
where n is the total number of items and r is the number of items chosen.
Now, let's solve each part of the question:
A. How many different committees can be formed?
Here, we need to select a committee of 4 reporters from a pool of 28. Since the order does not matter, we need to calculate a combination:
C(28, 4) = 28! / (4!(28-4)!)
Simplifying the expression:
C(28, 4) = 28! / (4!24!)
Using factorials:
C(28, 4) = (28 x 27 x 26 x 25) / (4 x 3 x 2 x 1)
Calculating:
C(28, 4) = 20475
Therefore, there are 20,475 different committees that can be formed.
B. How many different committees contain 3 men?
Here, we need to select 3 men from a pool of 16 men, and 1 additional reporter (regardless of gender) from the rest of the pool (28-1 = 27). Again, we will use combinations:
C(16, 3) = 16! / (3!(16-3)!)
Simplifying the expression:
C(16, 3) = 16! / (3!13!)
Using factorials:
C(16, 3) = (16 x 15 x 14) / (3 x 2 x 1)
Calculating:
C(16, 3) = 560
Therefore, there are 560 different committees that contain 3 men.
C. How many different committees contain at least 2 men?
To find the number of committees with at least 2 men, we need to sum up the following cases:
1. Committees with exactly 2 men: C(16, 2) * C(12, 2)
2. Committees with exactly 3 men: C(16, 3) * C(12, 1)
3. Committees with all 4 men: C(16, 4)
Calculating each case:
1. C(16, 2) * C(12, 2) = (16! / (2!14!)) * (12! / (2!10!))
2. C(16, 3) * C(12, 1) = (16! / (3!13!)) * 12
3. C(16, 4) = 16! / (4!12!)
Summing up these cases will give us the total number of committees with at least 2 men.
D. How many different committees contain at most 3 men?
To find the number of committees with at most 3 men, we need to sum up the following cases:
1. Committees with no men: C(12, 4)
2. Committees with exactly 1 man: C(16, 1) * C(12, 3)
3. Committees with exactly 2 men: C(16, 2) * C(12, 2)
4. Committees with exactly 3 men: C(16, 3) * C(12, 1)
Calculating each case and summing them up will give us the total number of committees with at most 3 men.