Question

In how many ways can the letters of the word "PERMUTATIONS" be arranged so that there are always 4 letters between 'P' and 'S'?

Permutations

Answer 1

`25,401,600`

Explanation:

There are 12 letters in "Permutations" - 1 P and 1 S (we'll work with the other letters in a minute).

We can put:

the P in place 1 and S in place 6
the P in place 2 and S in place 7
the P in place 3 and S in place 8
and so on, for a total of 7 placements.

We can also reverse the order of P and S, thus doubling the number of placements to 14.

Ok - so we have `14xx"ordering the rest of the letters"`

We have 10 letters to deal with that can go in the remaining 10 spots. Of the 10 letters, 9 are unique - there are two T's. If the 10 letters were all unique, we'd simply say that their ordering would equal `10!`:

• there being 10 choices of what to put in the first open spot
• then 9 remaining choices in the second open slot
• then 8...
• and so on

which gives `10xx9xx8xx7xx6xx5xx4xx3xx2xx1=10!`

But we need to divide by the number of ways we can order the T's (which is `2!`) to get rid of double counting. That gives:

`(10!)/(2!)` ways to order the remaining 10 letters.

All told, we have:

`14xx(10!)/(2!)=14xx(10!)/2=7xx10! = 7xx 3,628,800=25,401,600`