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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 $S$ ?
(i) words start with $P$ and end with $S$
(ii) vowels are all together,
(iii) there are always 4 letters between $P$ and $S$ ?
Solution:
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Verified Answer
(i) Letters between P and S are ERMUTATION. These 10 letters having $\mathrm{T}$ two times, these letters can be arranged in $\frac{10 !}{2 !}$ ways $=1814400$ ways
(ii) There are 12 letters in the word PERMUTATIONS. which have T two times.
Now the vowels a, e, i, o, u are together Let it be considered in one block. The letters of vowels can be arranged in 5 ! ways Thus, there are 7 letters and 1 block of vowel with $T$ two times
$\therefore$ Number of arrangements $=\frac{8 !}{2 !} \times 5 !=2419200$
(iii) There are 12 letters to be arranged in 12 place [1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11] [12] there are 12 letters are to be filled in 12 place. $P$ may be filled up at place No. 1, 2, 3, 4, 5, 6, 7 and consequently S may be filled up at place no. $6,7,8,9,10,11,12$. leaving four places between. Now $\mathrm{P}$ and $\mathrm{S}$ may be filled up in 7 ways.
Similarly, $S$ and $P$ may be filled up in 7 ways
Thus, $P$ and $S$ or $S$ and $P$ can be filled up in $7+7$ $=14$ ways
Now, the remaining 10 places can be filled by $=\frac{10 !}{2 !}$ ways
$\therefore \quad$ No of ways in which 4 letters occur between $P$ and $S$
$=\frac{10 !}{2 !} \times 14=\frac{3628800}{2} \times 14=25401600$
(ii) There are 12 letters in the word PERMUTATIONS. which have T two times.
Now the vowels a, e, i, o, u are together Let it be considered in one block. The letters of vowels can be arranged in 5 ! ways Thus, there are 7 letters and 1 block of vowel with $T$ two times
$\therefore$ Number of arrangements $=\frac{8 !}{2 !} \times 5 !=2419200$
(iii) There are 12 letters to be arranged in 12 place [1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11] [12] there are 12 letters are to be filled in 12 place. $P$ may be filled up at place No. 1, 2, 3, 4, 5, 6, 7 and consequently S may be filled up at place no. $6,7,8,9,10,11,12$. leaving four places between. Now $\mathrm{P}$ and $\mathrm{S}$ may be filled up in 7 ways.
Similarly, $S$ and $P$ may be filled up in 7 ways
Thus, $P$ and $S$ or $S$ and $P$ can be filled up in $7+7$ $=14$ ways
Now, the remaining 10 places can be filled by $=\frac{10 !}{2 !}$ ways
$\therefore \quad$ No of ways in which 4 letters occur between $P$ and $S$
$=\frac{10 !}{2 !} \times 14=\frac{3628800}{2} \times 14=25401600$
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