I have read and understood answer provided hereAssume that there is $30$ letters, $5$ of which are vowels.
My Result: There are $\binom{8}{3}$ combinations where you can write vowels. Each combination contains $5^3$ variations, then we can have $30^5$ variations of other letters on remaining places. I have multiplied all that
Why is my result wrong? I have understood the textbook result!
$\endgroup$2 Answers
$\begingroup$You are overcounting certain words.
For example, you are counting the word $aaaaaaaa$ a total of $\binom{8}{3}$ times.
The answer should be:
$\binom{8}{3}5^3 25^5+\binom{8}{4}5^4 25^4 + \binom{8}{5}5^5 25^5 + \binom{8}{6}5^6 25^2 + \binom{8}{1}5^7 25 + \binom{8}{8}5^8=30^8-\binom{8}{2}5^2 25^6 - \binom{8}{1}5^1 25 ^7 - \binom{8}{8}25^8$
$\endgroup$ $\begingroup$With a least $3$ vowels you can make words of length $8$ having $3,4,5,6,7$ and $8$ vowels. First start by counting words with exactly $3$ vowels and expand the situation to $4$ vowels and so on.
This can be summarized as: $$\sum_{k=3}^{k\leq8}\binom{8}{k}5^k25^{8-k}$$.
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