I am trying to formulate a formula for the following combinatorics problem:

Let's say I have $2$ letters $I$ and $L$ . $I$ is mandatory, meaning that it should appear in each permutation, $L$ is not. So i can have the following permutations:

$$|(I\text{ }L), (L\text{ }I), (I)| = \text{3 permutations}$$

For $3$ letters $I, L, M$ so that $I$ is mandatory and the other $2$ are optional, I have:

$$ |(\text{I L M}), (\text{I M L}), (\text{L I M}), \\ (\text{L M I}), (\text{M I L}), (\text{M L I}), \\ (\text{I L}), (\text{I M}), (\text{L I}), \\ (M I), (I)| = \text{11 permutations} $$

For $n = 4$ letters one of which is a mandatory letter $I$ which should be in each permutation, I have:

$$4! + (4 - 1)(3!) + (4 - 1)(2!) + 1 = \text{49 permutations}$$

So I ended up writing the following formula for any $n \geq 3$ when one letter of the $n$ letters (assuming $I$) is mandatory (each letter is different from the other):

$$n! + \left [ (n - 1) \cdot \sum_{i = 2}^{n - 1} i! \right ] + 1$$

Now, how can I generalise this formula when the number of mandatory letters of $n$ increases from $1$ to $k$ ?

  • $\begingroup$ In your last formula, there should be no $ k $, and you can incorporate the outsiders by having $i $ start at $1 $ and go to $ n $. $\endgroup$ – Arnaud Mortier Apr 6 '18 at 22:18
  • $\begingroup$ I am sorry, please check my edit $\endgroup$ – user3019105 Apr 6 '18 at 22:20
  • $\begingroup$ It does look better, but I think it is wrong - if I understand the problem correctly. Don't you agree with the general formula that I gave below? $\endgroup$ – Arnaud Mortier Apr 6 '18 at 22:25
  • $\begingroup$ I am trying to understand your formula, I have to revise combinatorics a little bit. Why do you think that the formula I wrote for $n \geq 3$ and $k = 1$ is wrong? $\endgroup$ – user3019105 Apr 7 '18 at 8:37
  • $\begingroup$ Just write down $ n=5 $ explicitly, you will see that your formula fails at the middle term. Your factor $ n-1 $ is just a special case of a binomial coefficient. $\endgroup$ – Arnaud Mortier Apr 7 '18 at 10:44

You have $ n-k \choose i $ ways to choose which optional letters you take, once you have decided to take $i$ of them.

Then $(k+i)! $ ways to order the letters that you have.

Therefore the answer is $$\sum_{i=0}^{n-k}{n-k \choose i}(k+i)! $$

  • $\begingroup$ Could you please provide a walkthrough on how to elaborate the formula you wrote? $\endgroup$ – user3019105 Apr 6 '18 at 22:22
  • $\begingroup$ You have to add the number of words containing $i$ optionals, over all $ i $. For fixed $ i $ you have $2 $ choices to make, which $ i $ letters you take and what is the order of your $ k+i $ letters in the end. The number of options for choice $2 $ is independent of the first choice made, therefore you multiply (fundamental principle of counting). $\endgroup$ – Arnaud Mortier Apr 6 '18 at 22:33

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