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Let's say I have the letters A, B, C, D. I want to know the number of combinations in the space _, _, _, or 3, where repetitions are allowed. For this, I can apply the formula C=$n ^ x$. In the example would be C=$4 ^ 3$=64.

My question is: How do I calculate this, but if only one letter can be repeated? Let's just say "A" can be repeated.

For instance:

A, A, A

A, B, A

C, A, A

D, A, B (This one does not repeat any, but is part of the result)

...

...

Thanks!

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    Have you thought about the problem? Please post your thoughts. – 2017-02-26
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    Breaking into cases based on how many of the repeatable letter are used seems like a good idea to me. If three $A$'s are used, must be AAA so one outcome. If two $A$'s are used, pick what the remaining letter is and pick where it is, $3\cdot 3=9$ outcomes (*specifically AAB,AAC,AAD,ABA,ACA,ADA,BAA,CAA,DAA*). Continue by counting how many if one $A$ is used and if no $A$'s are used and adding the results. – 2017-02-26

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It's helpful if you have more complicated problems of this type to know something about how binomial coefficients can represent choices between a set of items, but this problem is feasible without that.

If $A$ is not repeated (which includes not even present) there are $4\cdot 3\cdot 2 = 24$ options for the three places (since no repeats means you cannot reuse earlier choices).

If there are two $A$s, the non-$A$ space can go in $3$ places, and be any of the other $3$ letters, giving $3\cdot 3 = 9 $ options.

Finally there is only $1$ arrangement for all $A$s.

This gives $24+9+1=34$ options overall when only $A$ can be repeated.