Why should 10 letters be arranged? Aren't 4 letters between P and S, hence they need to be arranged separately. We should arrange only (12-6) + 1 (set of six P...S) out of which 2 are T's ---> (7!/2!)*(4!)=7!/12 OR 420 ways only

Anurag@Gurome wrote:There are 12 letters in the word PERMUTATIONS, where T occurs 2 times.

(3) There are always 4 letters between P and S:
Here P and S are fixed, so 10 letters are left, out of which 2 are T's
So, 10 letters out of which 2 are T's can be arranged in 10!/2! = 1814400 ways
Now letters P and S can be arranged so that there are 4 letters between them, which can be done in 2*7 = 14 ways
Required no. of ways = 1814400 * 14 = 25401600

ah, thanks - got it

Anurag@Gurome wrote:

Night reader wrote:Why should 10 letters be arranged? Aren't 4 letters between P and S, hence their need to be arranged separately, and we should arrange only (12-6) + 1 (set of six A...S) out of which 2 are T's ---> (7!/2!)*(4!)=6!/12 OR 420 ways only

Because when we fix P and S, then we are left with 10 letters, out of which any 4 letters can be there between P and S (we don't know these 4 letters, they can be any letter except P and S)

Vinooozz wrote:Dear friends,
I have a permutation sum copied below. I am not able to solve the iii)
Please help me resolve this with explanation

In how many ways can the letters of the word PERMUTATIONS be arranged if the
(i) words start with P and end with S, (ii) vowels are all together,
(iii) there are always 4 letters between P and S?

Thanks & Regards,
Vinu

The word PERMUTATIONS has 12 letters to it not all distinct, it has 2 Ts to it as well.

(i) PERMUTATIONS itself is a word that starts with P and ends with S. The remaining 10 letters need arrangements only, and 10 letters with only T occurring twice may be arranged in [spoiler]10! /2![/spoiler] number of ways. This is cumbersome to solve hence GMAT won't expect you to solve it fully.

(ii) Since PERMUTATIONS has E, U, A, I, and O all 5 vowels to it, and for those vowels to occur together we first need to take them out from PERMUTATIONS, and put them together in a bracket like (EUAIO) and consider this bracket as a single element now so that we now need to permute PRMTTNS(EUAIO) i.e. 8 elements only with only T occurring twice; never forget that the elements in the bracket (EUAIO) can further be permuted within themselves in 5! number of ways. Hence, the arrangements when vowels are all together are [spoiler](8! × 5!)/ 2![/spoiler], which again is cumbersome to solve hence GMAT won't expect you to solve it fully.

(iii) Consider PERMUSTATION as one case of this kind and SERMUPTATION as another where there are always 4 letters between P and S. We have 7 allocations possible to fix P and S in that way PERMUSTATION, and 7 allocations possible to fix S and P in that way SERMUPTATION. So, basically there are 14 ways to fix P and S, whereas the remaining 10 letters with only T occurring twice may be arranged in 10! /2! number of ways. Hence, the total arrangements in which there are always 4 letters between P and S is [spoiler]14 × 10! /2![/spoiler].

Thank you Anurag

Thank you so much. I was calculating that there are 4! numbers between p(s) and s(p) every time. Thank you I hope I now understand it. Thank you so much.