How many names exist

Question

Names:

Need-

• Name must start with J

• 4-5 letters long

• Exactly one z in name

How many names exist that meet all of these restrictions?

The name can be anything as long as it meets the above bullet points. ie Jzaa is fine

Really confused how to go about this problem was thinking of using n!/k!

Would really appreciate any help about how to go about this

Answer 1

Suppose the name is 4 letters long. Then it can be (J,z,[a-y], [a-y]) or (J, [a-y], z, [a-y]) or (J, [a-y], [a-y], z) where there are 25 choices for each of the [a-y] fields. So the number of possible 4-letter names 0s $3 \times 25 \times 25 = 1875$.

Use the same method to count the number of possible 5-letter names.

Answer 2

For a four-letter name, there are $\binom{3}{1}=3$ locations for the $z$. For each such location, the other two positions can be any of $a-y$, so $25^2$. Thus there are $3\cdot 25^2$ four-letter names. Similarly, for a five-letter name, there are $\binom{4}{1}=4$ locations for the $z$. For each such location, the other three positions can be any of $a-y$, so $25^3$. Thus there are $4\cdot 25^3$ four-letter names. So the total number is $$3\cdot 25^2 + 4\cdot 25^3 = 64375.$$

Answer 3

You have two choices as to whether the name is $4$ or $5$ letters long. Each is a different problem and the answer is the sum of the the two problems.

You have no choice on the first letter. It must be a "J".

You have a choice with the "z". The choice is what position the "z" belongs in. For the four-letter word there are $3$ choices and for the five letter word there are $4$.

For the remaining $2$ or $3$ numbers there are $25$ choices; the can be any letter except $z$.

So multiply the choices together and add the answer to the two different problems and you get.

$3*25^2 + 4*25^3$.