Permutations question: An experiment is to form "words" using all of the letters in pprmmtsss. What is the probability of a "word" having a p next to an m?

Question

Permutations question: An   experiment   is   to   form   "words"   using   all   of   the   letters   in   pprmmtsss.       What   is   the   probability   of   a   "word"   having   a   p   next   to   an   m?

anonymous · Answer

The total number of words would be 9!/(2! * 2! * 3!) = 15120

ganeshie8 · Answer

tie one pair of `pm` together

anonymous · Answer

There are two ways to to arrange p and m as either pm and mp. If we fix one m and one p as a single mp, then we can find the number of words with (mp), prmtsss, which would be 2*8*7!/3! = 13440. I believe there are 13440 words with at least 1 p next to 1 m, however this number includes duplicates.

anonymous · Answer

Using inclusion and exclusion principle, I will attempt to find the number of duplicates and then subtract from 13440.

anonymous · Answer

If we tie both the first m and p together as mp and the second m and p together as mp we will have (mp), (mp), r, t, s, s, s. There will be 2 * 2 * 7C2 * 5! / (2! * 3!) = 1680 duplicates.

anonymous · Answer

Using inclusion exclusion, the number of words with p next to m would be 13440 - 1680 = 11760, so the probability will be 11760/15120 = 7/9

ganeshie8 · Answer

Looks good !

anonymous · Answer

Thanks! @ganeshie8