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Simple combination math confusion

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Riffz
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Simple combination math confusion [#permalink] New post   25 Mar 2016, 07:33
I have a simple problem that needs a little explanation. The problem is as follows;

In how many combinations can we choose 2 students out of 10 if each student is needed to fill a different roll in the student's council?
(A) 110
(B) 45
(C) 55
(D) 90
(E) 100

Now if I use combination formula nCr = n!/r!(n-r)!, then the answer comes 45 (correct me if I am wrong), but where I got this problem says the answer is 90. The explanation is, "For the first roll there are 10 free student, for the second roll there are only 9 left. Therefore we have (10x9)= 90 combinations total."

Can anyone explain, which one is correct?



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Re: Simple combination math confusion [#permalink] New post   25 Mar 2016, 07:45
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Riffz wrote:
I have a simple problem that needs a little explanation. The problem is as follows;

In how many combinations can we choose 2 students out of 10 if each student is needed to fill a different roll in the student's council?
(A) 110
(B) 45
(C) 55
(D) 90
(E) 100

Now if I use combination formula nCr = n!/r!(n-r)!, then the answer comes 45 (correct me if I am wrong), but where I got this problem says the answer is 90. The explanation is, "For the first roll there are 10 free student, for the second roll there are only 9 left. Therefore we have (10x9)= 90 combinations total."

Can anyone explain, which one is correct?


Hi,
the Q here is that of permutation, where arrangement/order is important..
nPr=n!/(n-r)! or nCr*r!..

now lets do this Q by combinations.
you have correctly found that 45 ways are there to choose 2 out of 10..
But it is also given that each role is different..
so the choosen two can fill up these two different roles in 2! ways
so total ways = 45*2!=90
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Re: Simple combination math confusion [#permalink] New post   25 Mar 2016, 07:50
Thanks. That cleared my confusion ^_^

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Re: Simple combination math confusion [#permalink] New post   25 Mar 2016, 07:53
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Riffz wrote:
I have a simple problem that needs a little explanation. The problem is as follows;

In how many combinations can we choose 2 students out of 10 if each student is needed to fill a different roll in the student's council?
(A) 110
(B) 45
(C) 55
(D) 90
(E) 100

Now if I use combination formula nCr = n!/r!(n-r)!, then the answer comes 45 (correct me if I am wrong), but where I got this problem says the answer is 90. The explanation is, "For the first roll there are 10 free student, for the second roll there are only 9 left. Therefore we have (10x9)= 90 combinations total."

Can anyone explain, which one is correct?


Please follow the rules: rules-for-posting-please-read-this-before-posting-133935.html Thank you.

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Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
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Re: Simple combination math confusion [#permalink]
25 Mar 2016, 07:53
Moderator:
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