How permutation order matters here?

Question

How many ways are there to select a first-prize winner, a second-prize winner, and a third-prize winner from 100 different people who have entered a contest ?

This question is easy to solve as 1st prize can be given in 100 ways, then 2nd prize can be given in 99 ways and similarly, 3rd prize can be given in 98 ways, so total ways are => 100 * 99 * 98 => 970200 ways

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But 100 * 99 * 98 = P(100,3) and since it is a permutation. ordering must be there but I don't see any ordering here.

Can anyone explain that what am I missing ?

Answer 1

The order is there because you have reasoned that there are:

• 100 options for the first prize,
• 99 options for the second prize,
• 98 options for the third prize.

The order in which you then calculate the product of these three numbers is of course, due to commutativity, not important. But their values result from taking the order into account.

Without taking order into account, you would simply need to pick 3 people out of 100 and that's a combination of 3 out of 100: $${100 \choose 3} = 161\,700$$ It's no coincidence that you find precisely $\tfrac{1}{6}$th of the answer with the permutation because there are $3!=6$ ways to order these 3 randomly chosen people into the three prize winning positions.

Answer 2

In your calculation $P(100,3)=100\times99\times98$, you have written 100 possibilities for the first prize winner, 99 for second prize (as first prize winner can't also get the second prize simultaneously) etc. Suppose one of the counted possibilities was A as first prize winner, B the second prize winner and C the third prize winner.

As our count for first prize possibilities was 100 so B would also be counted as first prize winner (as B is part of the 100). Once first prize winner is fixed as B, for second prize you have written 99 possibilities, which is everything except the first prize winner B; this means A as second prize winner, and C as third prize winner will also be accounted. And so on. So from ABC we could see BAC etc. This takes care of all permutations.

Is it clear now?