i count the number of committees including Michel as following
let the committees be structured as : A B C
Order is not important
so, Michael comes in place of A
Place B can be taken by any of the 5 people left
Place C can be taken by any of the 4 people left
so possible committees including Michael = 1 x 5 x 4=20
but alternatively,
A is taken by Michael so 1 way
rest of the two places can be taken in 5c2 ways
so number of committees including Michael = 1 x 5c2=10
i need to know which solution is right and why is it so?
let the committees be structured as : A B C
Order is not important
so, Michael comes in place of A
Place B can be taken by any of the 5 people left
Place C can be taken by any of the 4 people left
so possible committees including Michael = 1 x 5 x 4=20
but alternatively,
A is taken by Michael so 1 way
rest of the two places can be taken in 5c2 ways
so number of committees including Michael = 1 x 5c2=10
i need to know which solution is right and why is it so?
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