How many names exist

discrete mathematics

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

Best Answer

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.

Related Solutions

Graph Theory – Showing that No Hamilton Circuit Exists

Suppose that $v$ is a vertex of degree $2$. If $G$ has a Hamilton circuit, that circuit must pass through $v$, so it must include both of the edges incident at $v$. In particular, if the configuration

              o----o----o  
              u    v    w

occurs anywhere in $G$, it must be part of any Hamilton circuit in $G$. Such a circuit must be a circuit, so it must include some other path from $u$ to $w$ as well. This can be useful in proving that a graph does not have a Hamilton circuit: if there is no other path between $u$ and $w$ in $G$, then $G$ cannot have a Hamilton circuit.

[Math] How many ways can six of the letters of the word ALGORITHM be selected and written in a row if the first letter must be A

We really need to choose $5$ non-A letters from the $8$ non-A's available, and put them in a row after the A.

The first letter (after the obligatory A) can be chosen in $8$ ways.

For each of these ways, the second letter can be chosen in $7$ ways.

For each choice of first and second, there are $6$ ways to choose the third, and then $5$ ways to choose the fourth, and $4$ ways to choose the fifth, for a total of $(8)(7)(6)(5)(4)$.

Another way: We can choose the $5$ letters from the $8$ available in $\binom{8}{5}$ ways. For every choice, we can line up the $5$ chosen letters in $5!$ ways, for a total of $\binom{8}{5}5!$.

Another way: We do not find the $\text{P}(n,r)$ notation particularly useful. But since it was mentioned in the post, we want to make a $5$-letter word (to append to A) from the $8$ non-A's. There are $\text{P}(8,5)$ ways to do this.

Best Answer

Related Solutions

Graph Theory – Showing that No Hamilton Circuit Exists

[Math] How many ways can six of the letters of the word ALGORITHM be selected and written in a row if the first letter must be A

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