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Here is the question:

We consider the Latin alphabet with $26$ letters, of which $5$ are vowels. How many words can we form that start with b and contain c and have $2$ vowels and $3$ consonants in total?

My answer:

We already have $2$ consonants so we need $1$ consonant and $3$ vowels, we have $4$ cases:

case one: b c _ _ _ case two: b _ c _ _ case three: b _ _ c _ case four: b _ _ _ c

In each case the vowels can be arranged in $C(3,2)$ ways so in total $4 \cdot C(3,2)$ ways and the consonants can be arranged in $C(3,1)$ ways so in total $4 \cdot C(3,1)$

Final answer: $4 \cdot C(3,2)+4 \cdot C(3,1)$

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  • $\begingroup$ You also have to choose which consonants fill the positions where you place the consonants and which vowels fill the positions where you place the vowels. $\endgroup$ – N. F. Taussig Oct 10 '20 at 9:09
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Your answer seems far too low…

There are $\binom42=6$ ways to arrange the two vowels and two remaining consonants, the latter of which includes the mandatory c. There are $20+20+1=41$ ways to choose the consonants: $c?,?c,cc$ where $?$ is any consonant apart from $c$. Then there are $5^2=25$ ways to choose the vowels, since they are unrestricted.

All in all, there are $6×41×25=6150$ admissible words.

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