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How many 10-letter words (they don't have to be actual words) can you make using 3 A's, 4 B's and the rest could be any letter from a 24-letter alphabet?

I'm trying to solve this problem, and my thoughts as for now are:

There are $10!9!8!$ ways to put the A's, then for every way there are $7!6!5!4!$ ways to put the B's, and then? I think there are so many special cases, but I don't quite know how to work this out.

Any thoughts?

Thanks in advance.

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3 Answers

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First choose $3$ places for $A$ and $4$ places for B.

Number of ways to do that is $\displaystyle {10 \choose 3} {7 \choose 4}$

Now in remaining $3$ places, put any of the $24$ letters in $24^3$ ways (considering repetitions are allowed).

So total number of words that can be made is $ \ \displaystyle \frac{10!}{3! \ 3! \ 4!} \cdot 24^3$

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Requirement: $10$ letter words.

$7$ letters are chosen. Ways to arrange them: $$\cfrac{7!}{3!4!}$$

Note: We divide by $4!3!$ because there are two groups of identical letters ($3,4$ in numbers, i.e. $3$ A's and $4$ B's).

A, B are to be placed in $7$ of the $10$ places. Therefore, ways to select $10$ places: $10C7$

Remaining $3$ can be any of the $24$ alphabets.

$\implies 24^3 $

[Because repetition must be allowed if they haven't explicitly mentioned that it isn't.]

Finally, by multiplication principle:

Answer: $$\cfrac{7!}{3!4!} \times 24^3 \times 10C7$$

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Out of those $10$ places, $3$ will be taken up by As. That's choosing $3$ places out of the $10$ available places: $$ {}^{10}C_3 \text{ ways } $$

Now, since $3$ places are already consumed, you've got $7$ places left, out of which $4$ would be taken up by Bs: $$ {}^{7}C_4 \text{ ways } $$

You may allot places to Bs first and then As and it'd still be the same.

After allotting positions for As and Bs you're left with $3$ positions which could be filled by $24$ remaining alphabets, assuming repetitions are allowed: $$ 24^3 \text{ ways } $$

So the total number of letters that could be made would be:$$ {}^{10}C_3 \times {}^{7}C_4 \times 24^3\text{ words} $$

If repetitions are not allowed in choosing those $3$ letters out of the $24$ alphabets, then that could be done in $$ {}^{24}P_3 \text{ ways} $$

So the total number of letters that could be made would be:$$ {}^{10}C_3 \times {}^{7}C_4 \times {}^{24}P_3\text{ words} $$

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