# How many possible length-8 strings

I'm working on what I think is a very simple problem, but I think I'm approaching it incorrectly.

Background: String length = 8, consists of X, Y, Z (must have one and only one occurrence each) and the digits 0 through 9 (can have any number of occurrences each).

I think I have the right idea for the first part? It asks for the total possibilities. So I made a table with each digit and the possible positions of that digit: \begin{array}{|c|c|} \hline Digit&Possible\ Positions \\ \hline X& 8 \\ \hline Y& 7 \\ \hline Z& 6 \\ \hline Numbers(10)& 5\\ \hline \end{array}

So then the possibilities are $8 * 7 * 6 * 5^{10} = 3,281,250,000$, right? Or do I have the wrong idea?

The second part I found a lot more difficult. It asked how many possibilities if no two letters can be consecutive.

The "X" has 8 possible positions. But after, that, the letters get harder: If my string is X _ _ _ _ _ _ _ , then there are 6 places where Y could go. But, if the string is _ X _ _ _ _ _ _, then there are only 5 places where Y could go.

\begin{array}{|c|c|} \hline Digit&Possible\ Positions \\ \hline X& 8 \\ \hline Y& *** \\ \hline Z& *** \\ \hline Numbers(10)& 5\\ \hline \end{array}

I'm just getting started with this, so simple/intuitive explanations are most welcome! :)

So, for the second part you will always have $5$ digits ($d_1$ to $d_5$) with $6$ interstices:

$$\_\,d_1\,\_\,d_2\,\_\,d_3\,\_\,d_4\,\_\,d_5\,\_$$

there will be $10$ possible digits for each of $d_1$ to $d_5$, so that's $10^5$ ways to replace $d_1$ to $d_5$.

Then if you place letters X,Y and Z in the interstices so that there is only one letter per space we fulfil the non-consecutive letter requirement.

There are $6$ spaces to place X, then $5$ spaces to place Y then $4$ spaces to place Z, so we get:

$$6\cdot 5\cdot 4\cdot 10^5\tag{Answer}$$

The number of ways to choose where the $X$ goes is $8$. Given this there are $7$ choices for the $Y$, and given that, there are $6$ choices for the $Z$. In the remaining $5$ slots, there are $10$ choices for each of them (the $10$ digits). This is $$8 \cdot 7 \cdot 6 \cdot 10^5 = 33600000$$

You were correct except that you wrote $5^{10}$ instead of $10^5$.

• Thank you! If you or someone else has time, I also am looking for an explanation of the follow-up question: how many possibilities in which no two letters can be consecutive. Feb 21, 2018 at 20:43
• For that follow-up question you should use complementary counting - it's easier to count the number of possibilities in which some two letters are consecutive (or all three) than it is to count the possibilities in which none are consecutive. Feb 21, 2018 at 20:49

For the second part you can follow the stars and bars method. If you set it up for your problem you should get ${8-3+1 \choose 3} 3!$ for the letters and same as first part for the numbers.