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You have a total of $9$ people to choose from. Of these $9$ people you are supposed to create a group of $4$. How many different ways can the new group look?

This is my reasoning:

To the new group, the first person can be chosen in $9$ ways, the second in $8$ ways, the third in $7$ ways and the fourth in $6$. Thus there can be $9 * 8 * 7 * 6 = 3024$ different types of groups with $4$ people.

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    $\begingroup$ You have chosen them in order, i.e. you have chosen a first, second, third, and fourth person. You need to divide by the number of times you can rearrange the people you picked, which is $4\cdot 3 \cdot 2 \cdot 1=24$. $\endgroup$
    – jgon
    May 21 '15 at 17:58
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You are almost there. One problem with your way of thinking is that the same group will be counted multiple times. With $n=9$ and $k=4$ you have calculated $$ \frac{n!}{k!} = \frac{9!}{5!} $$ What you need is the binomial coefficient: $$ \binom{n}{k} = \frac{n!}{(n-k)!k!} = \binom{9}{5} = \frac{9!}{(9-5)!5!} $$ You get the extra $(n-k)! = (9-5)! = 4!$ because the same group will be counted that many times.

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