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Probability

by Mochad » Mon Apr 02, 2012 11:07 am
Hi,

In how many ways a delegation of 4 students can be selected from a group of 12 if 2 students refuse to be selected at the same time?

My solution was to calculate the nbr of ways 4 students can be selected from 12 (4C12 = 12! / 4!(12-4)! = 450 way) then subtracting it from the nbr of ways in which the 2 students (e.g. A & B) are selected together. The way I solved it was (2C12 = 12! / 2!(12-2)! = 66) but the solution was 2C10 and not 2C12?

Can anybody help me and explain why 2C10 and no 2C10?
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by klmehta03 » Mon Apr 02, 2012 11:45 am
Combinations with restrictions:
with no restrictions= 12C4= 495
with restrictions= (10C2*2C2)
subtraction= 495-45= 450
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by Bill@VeritasPrep » Mon Apr 02, 2012 12:20 pm
Mochad wrote:Hi,

In how many ways a delegation of 4 students can be selected from a group of 12 if 2 students refuse to be selected at the same time?

My solution was to calculate the nbr of ways 4 students can be selected from 12 (4C12 = 12! / 4!(12-4)! = 450 way) then subtracting it from the nbr of ways in which the 2 students (e.g. A & B) are selected together. The way I solved it was (2C12 = 12! / 2!(12-2)! = 66) but the solution was 2C10 and not 2C12?

Can anybody help me and explain why 2C10 and no 2C10?
If we're looking for delegations that include the 2 students, they already take up 2 of the 4 spots. We have 2 remaining spots to fill, and we have 10 students to choose from (since the 2 students who don't like each other are already included).
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by Mochad » Mon Apr 02, 2012 2:17 pm
Thank you Bill :) I forgot that we should select 4 and not 2 members!!

BTW what is the difficulty of this question? Is it medium?
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