How many 5 person committees chosen at random from a group consisting of 3 men, 5 women, and 2 children contain at least 1 woman?
A) 250
B) 251
C) 252
D) 275
E) 300
Could someone explain to me how to solve this problem using Brett's FCP method? Thanks!
How to solve using the FCP method?
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- sanju09
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Shorter way to answer for at least 1 woman is to take 'no woman' number of ways out from all possible ways of choosing 5 out of a total of 10 people.hypedaj wrote:How many 5 person committees chosen at random from a group consisting of 3 men, 5 women, and 2 children contain at least 1 woman?
A) 250
B) 251
C) 252
D) 275
E) 300
Could someone explain to me how to solve this problem using Brett's FCP method? Thanks!
Total First
Make 5 blanks
_ _ _ _ _
Now, since first person can be chosen in 10 number of ways, second person can be chosen in 9 number of ways, and so on, hence place those options in order, over those blanks
(10)(9)(8)(7)(6)
_ _ _ _ _
Since order doesn't matter in selection process and we've 5 blanks up there, therefore we must divide the above product by (5)(4)(3)(2)(1) in order to get rid of repeat selections.
(10)(9)(8)(7)(6)
_ _ _ _ _ = 252 ways.
(5) (4)(3)(2)(1)
No Woman Next
'No woman' means either man or child and there are exactly 5 persons left if we don't consider women, hence there is exactly 1 way of forming a 5-person committee out of 5 persons.
Finally
Number of 5-person committee containing at least 1 woman = [spoiler]252 - 1 = 251
Pick B[/spoiler].
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Sanjeev K Saxena
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www.manyagroup.com
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- ceilidh.erickson
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Whenever you see the language of "at least" in a combinatorics or probability, start by considering the ways that that particular outcome won't happen. As sanju mentioned, there is only one way to pick a committee with no women on it: the committee of 3 men and 2 children. (And we'll ignore for the moment that children make terrible committee members!).
So, we simply want to solve for the total number of possible committees, then subtract out that 1 instance that doesn't work.
A quicker way to solve for the total number of possible committees is by using factorials. If we want to choose 5 people out of 10, we can divide the factorial of the total pool of people by the factorial of the people we choose times the factorial of the people we don't. In other words:
10!/(5!*5!)
This simplifies to (10*9*8*7*6)/(5*4*3*2), which simplifies further to 2*9*2*7 = 252
252 - 1 = 251
So, we simply want to solve for the total number of possible committees, then subtract out that 1 instance that doesn't work.
A quicker way to solve for the total number of possible committees is by using factorials. If we want to choose 5 people out of 10, we can divide the factorial of the total pool of people by the factorial of the people we choose times the factorial of the people we don't. In other words:
10!/(5!*5!)
This simplifies to (10*9*8*7*6)/(5*4*3*2), which simplifies further to 2*9*2*7 = 252
252 - 1 = 251
Ceilidh Erickson
EdM in Mind, Brain, and Education
Harvard Graduate School of Education
EdM in Mind, Brain, and Education
Harvard Graduate School of Education