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Question: In how many ways can the first and second place be awarded to two persons from among 9 people.

my answer is = 9!/(9-2)!2!
             = 9!/7!2!
             = 36

however the suggested answer I found on the book is

  9!/7! which is 72

Please advise which one is correct and please advise why, in case the suggested answer is correct. Thanks.

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U have calculated the number of ways to choose a combination of two person from a group of 9. But in a specific combination, there are two different ways of arranging the places. So times 2 ur answer u get 72?

In other words, there are 36 ways to pick two person from a group of 9, but in each way of picking, you can arrange the two persons in two ways, eg, AB and BA, there are hence 72 ways in total for ur case.

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  • $\begingroup$ Yes, it is a combination, but the variation, which depends on the order. Number of varations = n!/(n-k)! = 9!/(9-2)!. $\endgroup$ – georg Jul 25 '14 at 11:42
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First choose a person to get the first place. There are $9$ candidates. Then choose one to get the second place. There are $8$ candidates left. This leads to $9\times 8=72$ possibilities. Things would have been different if there had been no distinction among the two persons picked out. Naming them $A$ and $B$ the possibilities $(A,B)$ and $(B,A)$ would stand for the same selection. The double counting must be repaired then by dividing by $2$ giving $36$ as outcome.

In general if $k$ persons are picked out and each of them gets a special position then there are $$n\times\left(n-1\right)\times\cdots\times\left(n-k+1\right)=\frac{n!}{\left(n-k\right)!}$$ possibilities.

If $k$ persons are picked out that do not get a special position then there are $$\binom{n}{k}=\frac{n!}{k!\left(n-k\right)!}$$ possibilities. Here factor $k!$ somehow repairs the 'more than once' counting.

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