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What is the number of ways to select ten distinct letters from the alphabet $\{a, b, c, \ldots, z\}$, if no two consecutive letters can be selected?

There are a couple of formulas that I could use, but I am not sure how to use them.

The number of combinations of a given length that use elements from a given set, allowing repetition and with no missing elements in other words, each element of the given set occurs at least once in each combination: $\binom{k-1}{n-1}$ where where $n$ is the number of elements in the given set and $k$ is the length of the combination.

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OR

The number of combinations of a given length that use elements from a given set, allowing repetition: $\binom{n+k-1}{n-1}=\binom{n+k-1}{k}$ where $n$ is number of elements in the given set and $k$ is the length of the combination.

Please help me understand what I should do and how to go about solving this question?

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Here is another ways of visualizing the problem. First, we arrange $16$ blue balls and $10$ green balls so that no two of the green balls are consecutive. Then we label the balls.

Line up $16$ blue balls, leaving spaces between successive balls and at the ends of the row. There are $17$ such spaces, $15$ between successive blue balls and two at the ends of the row.

$$\square \color{blue}{\bullet} \square \color{blue}{\bullet} \square \color{blue}{\bullet} \square \color{blue}{\bullet} \square \color{blue}{\bullet} \square \color{blue}{\bullet} \square \color{blue}{\bullet} \square \color{blue}{\bullet} \square \color{blue}{\bullet} \square \color{blue}{\bullet} \square \color{blue}{\bullet} \square \color{blue}{\bullet} \square \color{blue}{\bullet} \square \color{blue}{\bullet} \square \color{blue}{\bullet} \square \color{blue}{\bullet} \square$$

To ensure that no two of the ten green balls are adjacent, we choose $10$ of these $17$ spaces in which to insert a green ball, which can be done in

$$\binom{17}{10}$$

ways. Next, we label the balls from left to right. The letters on the green balls are the desired subset of the alphabet in which no two of the selected letters are consecutive. For instance,

$$\color{green}{\bullet} \color{blue}{\bullet} \color{green}{\bullet} \color{blue}{\bullet} \color{blue}{\bullet} \color{blue}{\bullet} \color{green}{\bullet} \color{blue}{\bullet} \color{blue}{\bullet} \color{green}{\bullet} \color{blue}{\bullet} \color{green}{\bullet} \color{blue}{\bullet} \color{green}{\bullet} \color{blue}{\bullet} \color{blue}{\bullet} \color{blue}{\bullet} \color{green}{\bullet} \color{blue}{\bullet} \color{green}{\bullet} \color{blue}{\bullet} \color{green}{\bullet} \color{blue}{\bullet} \color{green}{\bullet} \color{blue}{\bullet} \color{blue}{\bullet}$$

corresponds to the selection $\{A, C, G, J, L, N, R, T, V, X\}$.

A similar question was asked a while ago, where $r$ non-consecutive numbers had to be selected from $n$ numbers. Basically, the problem at hand can be converted to a stars-and-bars problem with $11$ slots and $10$ bars, where the bars represent the indices of the selected letters. The remaining 16 letters must be divided among the $11$ slots, with at least one letter in each of the $9$ slots in the middle. Once one letter is placed in each of these slots, there are $26 – 10 – 9 = 7$ letters left. As such, the number of ways to select $10$ non-consecutive letters from the alphabet equals:

$${7 + 10 \choose 10} = {17 \choose 10} = 19448$$

Having chosen 10 non consecutive letters from the alphabet, see how the gaps created by them will look like. There will be 10 gaps such that no two are adjacent and 16 letters. Thus the number of ways of selecting 10 non consecutive letters is same as putting 10 “gaps” in a sequence of 16 letters so that no two are consecutive. Thus we must choose from the 15 in-between gaps among the 16 letters and the two at the ends. Thus the number of ways is $\binom{17}{10}$

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