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
The OA is the option D.
What is the best approach to solving this PS question? Is there a fast and easy way? I'd be thankful for your help experts.
Four people are to be selected to receive a modeling
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Hello Vjesus12.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
The OA is the option D.
What is the best approach to solving this PS question? Is there a fast and easy way? I'd be thankful for your help experts.
Let's take a look at your question.
We have to select 2 men from 10 (10C2), one woman form 5 (5C2) and one child form 2 (2C1). Hence, the number of ways is $$10C\ 2=\frac{10!}{8!\cdot2!}=\frac{10\cdot9\cdot8!}{8!\cdot2}=\frac{90}{2}=45$$ $$5C\ 1=\frac{5!}{4!\cdot1!}=\frac{5\cdot4!}{4!}=5$$ $$2C\ 1=\frac{2!}{1!\cdot1!}=2$$
Hence, the final answer is 45*5*2=450.
Therefore, the correct answer is the option D.
I hope this answer may help you.
I'm available if you'd like a follow-up.
Regards.
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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
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
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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.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
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
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