Create a string of five letters using the letters: A, B, C, D, E, F, G, H, I, J, K, L, M.

a) How many words contain at least one A?

b) How many words contain exactly two A's?

For a) by my understanding is the total amount of combinations - the total amount of words without them. Which is what I believe to be $13^5 - 12^5$, however it feels as though it is too high/wrong

As for b), my train of thought is along the lines of:

$$\frac{13!}{2!\cdot 11!} \cdot \frac{12!}{3!\cdot 9!} \cdot \tfrac12$$

The first fraction is for the two A's, the second fraction is for the letters in the other three positions and the half is to get rid of the doubles in the set. ie AABCD and AABCD where the A's are swapped.


The (a) part is correct. For the (b) part, think that indeed you must choose the places for putting the two $A$'s: $\binom{5}{2}$. Then, each of the other three places can be ocuped for any of the remaining 12 letters: $12^3$. So, the (b) answer is $\binom{5}{2}12^3$.

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  • $\begingroup$ The English translation of restant is remaining. $\endgroup$ – N. F. Taussig Mar 16 '16 at 14:16

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