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In how many ways can you rearrange the letters A, B, C, D, E?

5! or ##5*4*3*2*1 = 120## ways

You have 5 letters in all and 5 spaces. Your first letter can be any of the 5. A, B, C, D, or E. Therefore you have 5 choices.

Your second letter can be any of the 5 original letters minus the one which took the first place. Thus you'll have only 4 choices and so on.

If you use combination for this problem, you can also look at it this way:

For the first place, you have 5 letters and 1 to select i.e. 5C1.

The second one will only have 4 letters but still 1 to select i.e. 4C1.

This will give you ##5C1 * 4C1 * 3C1 * 2C1 * 1C1## which is the same as ##5*4*3*2*1## or simply 5!

All would result in the same answer = 120.

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