PERMUTATIONS शब्द के अक्षरों को कितने तरीकों से व्यवस्थित किया जा सकता है, यदि
(i) चयनित शब्द का प्रारंभ P से तथा अंत S से होता है ।
(ii) चयनित शब्द में सभी स्वर एक साथ हैं ?
(iii) चयनित शब्द में P तथा S के मध्य सदैव 4 अक्षर हों ?

Explanation

We have letters P,E,R,M,U,(T,T),A,I,O,N,S.
Now between P and S there are four letters. We have following three cases.
Case I : Two of the four letters between P and S are TT.
In this case remaining two letters can be selected in `.^(8)C_(2)` ways. Suppose these letters are E and R.
So, we have to arrange (P, E, R, T, T,S), M,U,A,I,O,N.
`therefore` Number of arrangements in this case is
`.^(8)C_(2)xx7!xx2!xx(4!)/(2!)=3386880`
Case II : Four letters between P and S are distinct of which one letter is T.
In this case remaining three letters can be selected in `.^(8)C_(3)` ways. Suppose these letters are E, R, M.
So, we have to arrange (P,E, R, M, T, S), U, T, A, I, O, N.
`therefore` Number of arrangements in this case is
`.^(8)C_(3)xx7!xx2!xx4!=13547520`
Case III : Four letters between P and S are distinct other than T. In this case remaining four letters can be selected in `.^(8)C_(4)` ways.
Suppose these letters are E, R, M, U
So, we have to arrange (P, E, R, M, U, S),t, A, T, I, O, N
`therefore` Number of arrnagements in this case is
`.^(8)C_(4)xx(7!)/(2!)xx2!xx4!=8467200`
Hence, from all above cases, total number of ways are
3386880+13547520+8467200=25401600