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How many ways are there to make different words (not necessarily meaningful) by changing order of letters in the word $b_1b_1...b_1b_2b_2...b_2...b_mb_m...b_m$, where $b_1$ appears $k_1$ times, $b_2$ appears $k_2$ times, ... , $b_m$ appears $k_m$ times.

My answer is $$\sum_{k_1+k_2+...+k_m = n \\ 0\leq k_i \leq n}\quad\Large\frac{n!}{k_1!k_2!...k_m!}b_1^{k_1}b_2^{k_2}...b_m^{k_m}$$

Is this correct?

John Doe
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    Try out some small example to test your formula, with say one a, two b's, three c's – true blue anil Dec 30 '22 at 11:09
  • @trueblueanil would the answer be 60? – John Doe Dec 30 '22 at 11:23
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    What does it mean to raise $b_1$ to some power? – David P Dec 30 '22 at 12:09
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    If all the numbers would have been distinct, you would have $n!$ ways. If you make 4 of them the same, you can permute the 4 by fixing the others in $4!$ ways, so you're counting every case $4!$ times. So, there would be $n!/4!$ ways. Try to extend this to more groups of the same thing – D S Dec 30 '22 at 12:30
  • @DavidP It means permutations of $b_1$ identical letters? So for k=3, $b_1^k$ would be ways to make 3 letter words out of $\alpha,\beta,\gamma$, so 321 = 3! – John Doe Dec 30 '22 at 12:38
  • @trueblueanil Isn't that what the question is asking? – John Doe Dec 30 '22 at 12:58
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    I have written an answer to point out a possible source of your confusion. – true blue anil Dec 30 '22 at 13:12

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Your answer of $60$ corresponds to $\large\frac{6!}{1!2!3!}$, the first part of your formula, which represents the number of permutations of a $n$ letter word with $k_1$ of the first kind, $k_2$ of the second kind,.... $\Sigma k_i = n$

This represents the multinomial coefficient

What you have possibly mixed it up with is the Multinomial Theorem, which states

If events $E_1, E_2, ... E_k$ can occur with probabilities $p_1, p_2, ... p_k$ respectively, then the probability that events $E_1, E_2, ... E_k$ will occur $X_1, X_2, ... X_k$ times respectively is

$$\Large\frac{N!}{X_1!X_2! ... X_k!}p_1^{X_1}p_2^{X_2}... p_k^{X_k}$$

where $X_1 + X_2 + ... X_k = N$