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Q: Combination, Probability Question: Verification Request ( Answered 5 out of 5 stars,   2 Comments )
Subject: Combination, Probability Question: Verification Request
Category: Reference, Education and News > Homework Help
Asked by: unknownentity-ga
List Price: $5.00
Posted: 19 Nov 2006 14:22 PST
Expires: 19 Dec 2006 14:22 PST
Question ID: 784079
I calculated the answer to the following question to be:

Of the 100 people in the US Senate, 18 serve on the Foreign Relations
Committee.  How many ways are there to select Senate members for this
committee (assuming party affiliation is not a factor in selection)?

I used nCr=n!/(n-r)!r!


I am not sure if this is right, and just want some clarification.

Request for Question Clarification by mvguy-ga on 19 Nov 2006 17:37 PST
Would an answer to 14 significant digits be sufficient?

Clarification of Question by unknownentity-ga on 19 Nov 2006 17:40 PST
Most likely; I just need to use it to see where and if I went wrong with my math.
Subject: Re: Combination, Probability Question: Verification Request
Answered By: livioflores-ga on 19 Nov 2006 19:21 PST
Rated:5 out of 5 stars

This is a counting problem, specifically an unordered selection:
"Ordered Selection:
In general, the number of ordered selections of m items out of n items is:

The idea is that we cancel off all but the first m factors of the n! product.

Unordered Selection:
In general, to choose (unordered selection) m candidates from n, there are:
ways. We took the formula for ordered selections of m candidates from
n, and then divided by m! because each unordered selection was counted
as m! ordered selections.
From "HSE Basic counting - Wikibooks, collection of open-content textbooks":

The Unordered Selection is the selection that fits this problem, so
you have used the correct formula:
100!/(100-18)!18! = 100!/82!18! =
                  = 3.06645108029882083E+19 =
                  = 30664510802988208300

Your math for this problem is right.

For further references see:
"Permutations and Combinations":

"Permutations and Combinations" by Joe Sawada:

Search strategy:
"How many ways are there to select"
permutations combinations

I hope this heps you. Feel free to request for a clarification if you need it.

unknownentity-ga rated this answer:5 out of 5 stars
You're brilliant. :-)

Subject: Re: Combination, Probability Question: Verification Request
From: ansel001-ga on 19 Nov 2006 16:08 PST
I didn't multiply it out but your formula is correct.
Subject: Re: Combination, Probability Question: Verification Request
From: mvguy-ga on 19 Nov 2006 18:12 PST
I'm 95 percent that your formula is correct. If it is correct, your
multiplication is correct. The answer I get using your formula is
3.0664510802988 x 10^19, which is what you have. But I'm not certain
enough that you have the right formula to charge you for an answer.

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