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D I S C R E T E M A T H E M A T I C S I S R E A L L Y F U N

is there such that the words discrete, mathematics, is, really, fun do not appear consecutively.

I know that the way to remove repition here is to do the number of possible permutations Divided by the factor of the number of each repeated letter or

N!/(n_1!n_2!...n_n!)

And I can also use the inclusion exclusion to solve this. My question is for example if I'm trying to find the set of every possible appearance of DISCRETE for example, would this be 23!/2!?

spstephens
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1 Answers1

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D I S C R E T E M A T H E M A T I C S I S R E A L L Y F U N consists of

$\{E^4, A^3, I^3, S^3, T^3, C^2, L^2, M^2, R^2, D, F, H, N, U, Y\}$

So, yes, there is a total of $\frac{(4+3*4+2*4+6)!}{4!\,3!^4\,2!^4\,1!^6}=532995876358730104320000000$ distinct ways to arrange those letters.

Now to count the ways where the words "discrete", "mathematics", "is", "really", "fun", don't appear consecutively, we need to know ... what does that even mean?

Is it "all these words don't appear consecutively, in that order.": in which case there's only one way they can, so there are $532995876358730104319999999$ ways they don't.

Is it "all these words don't appear consecutively, in any order.": in which case there's $5!$ ways they can, so there are $532995876358730104319999880$ ways they don't.

Or is it something else? Please specify.

I'm sorry, the question is how many permutations of those letter are there such that none of those words appear as consecutive letters.

Right. Then Let $\mathcal D$ be the event that DISCRETE appears within the string, $\mathcal M$ the event that MATHEMATICS does, and so forth for $\cal I, R, F$ .

We note that the problem child is IS which can occur up to three times unless the letters are used in DISCRETE or MATHEMATICS.

  • $\lvert\mathcal D\rvert = \frac{(30-8+1)!}{3!\,2!^6}$
  • $\lvert\mathcal M\rvert = \frac{(30-10+1)!}{4!\,2!^5}$
  • $\lvert\mathcal R\rvert = \frac{(30-6+1)!}{3!^4\,2!^3}$
  • $\lvert\mathcal F\rvert = \frac{(30-3+1)!}{4!\,3!^4\,2!^4}$
  • $\lvert\mathcal I\rvert = \frac{(30-2+1)!}{4!\,3!^2\,2!^6}-\frac{(30-4+2)!}{2!\;4!\,3!^2\,2!^4}+\frac{(30-6+3)!}{3!\;4!\,3!^2\,2!^4}$
  • $\lvert \mathcal D\cap\mathcal M\rvert =\frac{(30-8-10+2)!}{2!}$
  • $\lvert\mathcal D\cap R\rvert = $ etc.
Graham Kemp
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