# Find The number of ways?

The number of ways in which all the integers from 1 to 36 (both inclusive) can be arranged such that no two multiples of 6 are adjacent is expressed as

$$m! x^n Pr$$ where m, n, r are distinct positive integers.

What is the sum m + n + r?

How i can achieve this? Thanks in advance.

EDIT: The formula is $$m!\times{^nP_r}$$ where $^nP_r = \frac{n!}{(n-r!)}$ is the number of $r$-permutations of $n$ or sequences without repetition of length $r$ chosen from an $n$-element set.

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What is $P$?  – draks ... Apr 17 '12 at 10:29
It stands for permutation.Like n P r. Its x^n P r. – vikiiii Apr 17 '12 at 10:35
And what is $x$? "The number of ways in which all the integers from 1 to 36 (both inclusive) can be arranged such that no two multiples of 6 are adjacent" is a constant number. Given such a number, how does one defines $m$, $x$, $n$ and $r$? – penartur Apr 17 '12 at 10:37
So I think the correct formula is $m!\times{}^nP_r$ ($m!\times{}^nP_r$) or $m!\times{}_nP_r$ ($m!\times{}^nP_r$) where ${}_nP_r=\frac{n!}{r!}$. – bgins Apr 17 '12 at 11:02

To find the number of ways, consider that the 30 numbers that are not multiples of 6 can first be arranged and then the remaining 6 can be arranged somewhere in the remaining 31 gaps. So the number is $30!\times ^{31}P_6$. Not sure how x fits into this (tbh looks like a typo).
EDIT: Looks like $x$ here meant multiplication. In that case m = 30, n = 31, r = 6. So m + n + r = 67.
Wow, you're great at telepathy! It seems that $x$ has been used as a replacement for $\times$ (so that it is not $x$ raised to $n$-th power but rather $P(6, 31)$). – penartur Apr 17 '12 at 10:44
Very nice! $nCr={n\choose r}=\frac{n!}{(n-r)!\,r!}$ ("$n$ choose $r$") and $nPr=\frac{n!}{(n-r)!}$ ("$n$ permute $r$") are alternate notations for choosing (unordered) and arranging (ordered) $r$ of $n$ objects; apparently some authors also superscript the $n$ as in ${}^nC_r$ and ${}^nP_r$. – bgins Apr 17 '12 at 10:59