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I need help with the following question:

"Given that a computer can only type letters A,B,C,D and E, how many ways can I type in 6 letters such that they must contain at least all of the different letters?

ie. AABCDE ABCDEA ABCDEB"

My professor said that the number of ways is: 6C4 * 5! because, 5! - There are 5! ways to arrange the letters 6C4 - Since 4 letters are going to be unique, out of the 6 letter positions, choose 4 positions to have these unique letters.

However, I had a different answer(which was wrong). Mine is: 6C5 * 5! * 5 I chose 5 out of 6 positions to include all ABCDE, and there are 5! ways of arranging them. As for the last position, it can include any of the 5 letters.

Could anyone explain to me why is my answer wrong?

Major thanks to all :3

user111400
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2 Answers2

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You can also try to solve it this way: ABCDE_ the only space left can be A,B,C,D or E, so it will be 5C1 way to fill up the empty space. And you will have total of 6 object with exactly 2 repeating.

Hence, (5C1*6!)/2! = 1800.

Cuiga
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Choose the first five positions to include ABCDE, and an ordering of them, say $$ \text{EBDCA_},$$ and then pick one of our five options for the last space, say A, so we get $$\text{EBDCAA}.$$

Alternately, we choose the first four and the last position to include ABCDE, and an ordering of them, say $$\text{EBDC_A},$$ and then pick one of our five options for the last space, say A, so we get $$\text{EBDCAA}.$$ Clearly, these are the same result, but your method counts them as though they are not.

It turns out that your answer is exactly double the correct answer. That is because, for any final arrangement of letters, your method counts two "different" ways to obtain it (as we saw above for one particular arrangement).

Cameron Buie
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