Hi VJesus12,

We're told that 4 people are to be selected to receive a modeling contract from a group of 10 men, 5 women and 2 children. We're asked for the number of possible ways in which 2 men, one woman and one child will receive a contract. You can answer this question by using the Combination Formula three times (although you really just have to use it once).

Since we're choosing 2 men, 1 woman and 1 child from each of their respective groups, we have three calculations to perform:

Men = choose 2 from a group of 10 = 10!/2!8! = (10)(9)/(2)(1) = 45 sets of 2 men
Women = choose 1 from a group of 5 = that's 5 options (you could also write this as 5!/1!4! = 5/1 = 5)
Children = choose 1 from a group of 2 = that's 2 options (you could also write this as 2!/1!1! = 2/1 = 2)

To get the total number of possibilities, we have to multiply those 3 outcomes: (45)(5)(2) = 450

Final Answer: D

GMAT assassins aren't born, they're made,
Rich

VJesus12 wrote:Four people are to be selected to receive a modeling contract from a group of 10 men, 5 women and 2 children. How many ways are possible in which 2 men, one woman and one child will receive a contract?

A. 90
B. 180
C. 225
D. 450
E. 900

We are given that 4 people are to be selected for a modeling contract. We must determine, from 10 men, 5 women, and 2 children, the number of ways that 2 men, 1 woman, and 1 child will receive a contract.

The number of ways that 2 men will receive a contract is 10C2 = (10 x 9)/2! = 45.

The number of ways that 1 woman will receive a contract is 5C1 = 5.

The number of ways that 1 child will receive a contact is 2C1 = 2.

Thus, the number of ways such that 2 men, 1 woman, and 1 child will receive a contract is:

45 x 5 x 2 = 450

Answer: D