# How many $5$-letter words can we form starting with b and containing c and having $2$ vowels and $3$ consonants?

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?

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

• 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. Oct 10, 2020 at 9:09

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.