# Thread:

1. ## Counting (Permutations/Combinations)

A group of 10 is going to be selected from a pool of 8 men and 8 women...
In how many ways ways can the selection be carried out if:
a) we choose 10 people at random?
b) there must be 5 men and 5 women?
c) there must be more women than men?

Also, if you have time, can you answer:
- How many strings of 6 decimaal digit.s have exactly three digits that are 4's?

I suck at this problems and help would be great! Thank you!

2. Originally Posted by achua7
A group of 10 is going to be selected from a pool of 8 men and 8 women...
In how many ways ways can the selection be carried out if:
a) we choose 10 people at random?
b) there must be 5 men and 5 women?
c) there must be more women than men?

Also, if you have time, can you answer:
- How many strings of 6 decimaal digit.s have exactly three digits that are 4's?

I suck at this problems and help would be great! Thank you!
You only need to know that there are $\binom{n}{k}$ ways to choose a k-element subset from an n-element set.

Thus you get
a) $\binom{10+10}{10}=\binom{20}{10}=184'756$

b) $\binom{10}{5}\cdot\binom{10}{5}=\binom{10}{5}^2=63 '504$

c) Let m be the number of men chosen, the corresponding number of women must then be 10-m; to satisfy the condition that m<10-m, m can only assume the values m=0, 1, 2, 3, or 4. Hence, by summing all these cases we get

$\sum_{m=0}^4\binom{10}{m}\cdot\binom{10}{10-m}=\sum_{m=0}^4\binom{10}{m}^2=60'626$

As to the number of strings of digits (not numbers) of length 6 that contain exactly 3 times the digit 4: you can choose the position of the 3 digits that are 4 in $\binom{6}{3}$ ways, and each of these can be combined with one of the $9^2$ ways to choose the remaining 2 digits, different from 4.
Thus you get $\binom{6}{3}\cdot 9^2=1'215$ possibilities.
Note that there is a difference depending on whether we are talking here of the number of strings of length 6 or of numbers with 6 digits, because in the latter case, the first digit would not be allowed to be 0.
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