Hi Dave,

First of all, you got the Combination and Permutation mixed up - but I guess that was the whole point of your post

I came up with this mnemonic device to keep these straight:

**P=Prizes (permutations)**
**C=Committee (combinations)**
Here's the full story:

[/SIZE]

**Permutations** ("

Prizes")[/SIZE]

Let's say you have 5 people who are running a race, and you want to know in how many ways three prizes (gold, silver, and bronze medal) could be awarded.

Any of the 5 could win the gold, any of the remaining 4 could win silver, and any of the other 3 could win bronze.

So you get 5 x 4 x 3 possibilities.

You can get this result using the permutation formula:

P(5,3) = 5! / (5-3)! = (5 x 4 x 3 x 2 x 1) / (2 x 1) = 60

Notice that the effect of the denominator in the formula is just to cancel out the last two terms from the numerator. That's why I prefer to think of it as simply an incomplete factorial where you multiply out only as many terms as you have selections. In this case, instead of the complete factorial of 5 x 4 x 3 x 2 x 1, you only have three prizes to be awarded, so you stop after the third term.

[/SIZE]

**Combinations** ("

Committee")[/SIZE]

Now think of the same 5 people, but your task this time is to form a committee of 3 (with no special roles, just equal members). You could do your selection in the same way as above, but you would find that some of these permutations give you the same committee. For example, it doesn't matter whether your selection is person A, then C, then D or whether it is C, then D, then A. So the straightforward selection process we used to award prizes for the race needs to be adapted a bit. We need to divide by the number of possible ways in which a particular committee could have been picked. This is simply the factorial of the number of committee members, in this case 3!

For example, consider these 6 permutations:

ACD

ADC

CAD

CDA

DAC

DCA

These selections all result in the same committee, so they are all equivalent to a single combination.

That's why the combination formula includes the division by n!, so

C(5,3) = P(5,3) / 3! = (5 x 4 x 3) / (3 x 2) = 10

Hope that clears it up. I completely agree with you that understanding the concept behind it is a much more solid approach than just memorizing the formulas. Once you understand the concept, you can reconstruct the formulas very quickly, even if you forget them under pressure. Also, if you can "map" the problem to one of the two scenarios above, you should have no problem picking the correct formula to use.

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