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25 Mar 2016, 07:33
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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?
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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25 Mar 2016, 07:45
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Riffz
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
