How many words can be formed using two letters "a", four letters "u" and 3 more? How many words can be formed using two letters "a", four letters "u" and 3 more?
My attempt: a combination could be "aauuuuxyz", where x, y or z can be chosen between 26 letters. I think the order is not relevant, because the first two "a" could be permuted, so I would think that our case is combination. But with the last three letters (x, y or z) the order is relevant. So it seems difficult to me to see how to do it.
 A: I think the best way is to use exponential generating functions,
find the coefficient of $x^9$ in
$9! (\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\frac{x^5}{5!})(\frac{x^4}{4!}+\frac{x^5}{ 5!}+\frac{x^6}{6!}+\frac{x^7}{7!})(x^0 +x +\frac{x^2}{2!}+\frac{x^3}{3!})^{24}$,
wolfram gives the answer  $18613092$

PS:
A bit of explanation on exponential g.f. use here

*

*Were all $9$ letters distinct, we would have $9!$ permutations, but they aren't


*The first term represents that there have to be two $a's$ which can be permuted in $2!$ ways, hence $x^2/2!$ and can be at most $5 a's$ (as there are already four $u's$) and so on.


*the last term represents zero to three from the remaining $24$ letters of the alphabet
A: There are $10$ possibilities for the three unspecified letters as $i$ a's, $j$ u's and $k$ not-a-or-u's. For each case the number of possibilities can be counted as $\binom9{2+i,4+j,k}24^k$ where a multinomial coefficient is used, which upon summing gives the answer as $18613092$.
A: The word shuold be like "aauuuu???" .
The possible cases of "???" are:

*

*$xyz$


*$uxy$ and $axy$


*$uux$ and $aax$ and $axx$ and $uxx$


*$uuu$ and $aaa$ and $uaa$ and $uua$


*$xxx$ and $xxy$
And then we can bombine the possible cases separately with the possible cases of "a"s and "u"s:

*

*$24 \times 23 \times 22 \times 3! \times C^2_9 \times C^4_7$


*$24 \times 23 \times 2! \times C^2_9 \times C^5_7 + 24 \times 23 \times 2! \times C^3_9 \times C^4_6$


*$24 \times C^6_9 \times C^2_3 + 24 \times C^4_9 \times C^4_5 + 24 \times C^5_9 \times C^2_4 + 24 \times C^4_9 \times C^3_5$


*$C^7_9 + C^5_9 + C^4_9 + C^6_9$


*$24 \times C^3_9 \times C^2_6 + 24 \times 23  \times 2 \times C^2_9 \times C^4_7 \times C^2_3$
Then we should add them all to get the final answer.
