Probabaility of having certain number of parents in a committee

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Homework Statement



A committee of 5 members is formed from 6 parents , 2 teachers, and a principal, In how many ways, can this committee be formed if the committee consists of not more than 4 parents?

my working is 1 parent + 2 parents + 3 parents +4 parents

which is (6C1 x 8C4) + (6C2 x 7C3) +(6C3x 6C2)+ (6C4 x 5C1)

but the ans given is only 120

Homework Equations





The Attempt at a Solution

 
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desmond iking said:

Homework Statement



A committee of 5 members is formed from 6 parents , 2 teachers, and a principal, In how many ways, can this committee be formed if the committee consists of not more than 4 parents?

my working is 1 parent + 2 parents + 3 parents +4 parents

which is (6C1 x 8C4) + (6C2 x 7C3) +(6C3x 6C2)+ (6C4 x 5C1)

but the ans given is only 120

Homework Equations





The Attempt at a Solution


Since there are only 3 non-parents your choices for number of parents are 2,3, and 4. Once you have counted the parents, the others must be chosen from the non-parents. Try again.
 
LCKurtz said:
Since there are only 3 non-parents your choices for number of parents are 2,3, and 4. Once you have counted the parents, the others must be chosen from the non-parents. Try again.

Does the identities of the parents and teachers matter? In other words, would a 3-parent committee having parents Smith, Jones and Chakravarty be counted separately from a 3-parent committee having parents Smith, Jones and Salam? Ditto for the teachers.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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