Question

In how many ways can the letters of the word 'PERMUTATIONS' be arranged if the
(i) words start with P and end with S and \[\]
(ii) Vowels are all together.\[\]
(iii) There are 4 letters between P and S. \[\]

Answer 1

Hint: We use the formula of arranging $n$ distinct objects as $n!$ and $n$ objects where $m$ objects repeats them by ${{p}_{1}},{{p}_{2}},...,{{p}_{m}}$ times as $\dfrac{n!}{{{p}_{1}}!{{p}_{2}}!...{{p}_{m}}!}$. In part(i) we fix P on first and S on ${{12}^{\text{th}}}$ position and then fill and arrange the rest 10 letters using permutation with repetition formula . In part(iii) We treat the vowels as a single letter and arrange all the letters using permutation with repetition formula. In part(iv) we multiply the number of ways we can place P and S 4 letters apart and the number of ways we can arrange the rest 10 letters. \[\]

Complete step-by-step solution:
We know that $n$ distinct objects can be arranged in particular in $n!$ ways and $n$ objects can be arranged with $m$ objects repeating themselves by ${{p}_{1}},{{p}_{2}},...,{{p}_{m}}$ times in $\dfrac{n!}{{{p}_{1}}!{{p}_{2}}!...{{p}_{m}}!}$ ways. \[\]
 We are given in the question the word ‘PERMUTATIONS’. We see that it has 12 letters and with only the letter T repeating itself by 2 times. So we have 12 positions to fill up the letters.
\[\begin{matrix}
   \_ & \_ & \_ & \_ & \_ & \_ & \_ & \_ & \_ & \_ & \_ & \_ \\
   1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 \\
\end{matrix}\]
(i) If the word has to start with P and has to end with S then the position of P is fixed on first position and position S is fixed on ${{12}^{\text{th}}}$ position.
\[\begin{matrix}
   P & \_ & \_ & \_ & \_ & \_ & \_ & \_ & \_ & \_ & \_ & S \\
   1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 \\
\end{matrix}\]
We can fill the rest 10 letters with only the letter T repeating itself 2 times in 10 positions between them. We use the permutation formula with repetitions and find required number of words with $n=10,{{p}_{1}}=2$ as
\[\dfrac{n!}{{{p}_{1}}!}=\dfrac{10!}{2!}=1814400\]
(ii) The vowels in the word ‘PERMUTATIONS’ are E, U, A, I , O. If the vowels come together then E, U, A, I , O will come together. Let us treat the vowels as a single letter unit EUAIO and the number of letters left is $12-5=7$. So the total number of letters with vowel unit EAIOU is $7+1=8$.
\[\begin{array}{*{20}{c}}
  {[E} &U &A &I &{O]} &\_&\_&\_&\_&\_&\_&\_ \\
  {[1}&1&1&1&{1]}&2&3&4&5&6&7&8
\end{array}\]

We can arrange rest $n=8$ letters with the letter T repeating itself ${{p}_{1}}=2$ times in $\dfrac{n!}{{{p}_{1}}!}=\dfrac{8!}{2!}$ ways but the 5 vowels within the unit can arrange themselves in $5!$. So by the rule of product, the total number of words is,
\[\dfrac{8!}{2!}\times 5!=2419200\]
(iii) If there are 4 letters in between P and S the position of S will be $x+5$ if we fix the position of P as $x$ or the position of P is $x+5$ if we fix position S as $x$ which we show on the table below.\[\]

Position of P	Position of S	Position of S	Position of P
1	6	1	6
2	7	2	7
3	8	3	8
4	9	4	9
5	10	5	10
6	11	6	11
7	12	7	12

So the number of ways we can fix P and S with 4 letter distance is 14.We can fill the rest $n=10$ letters with only the letter T repeating itself ${{p}_{1}}=2$ times in rest 10 positions in $\dfrac{n!}{{{p}_{1}}!}=\dfrac{10!}{2!}$ ways. We use the rule of product and find the required number of words as
\[\dfrac{10!}{21}\times 14=10!\times 7=25401600\]

Note: We note that the positions are fixed here if they would not have we would have consider derangement ${{D}_{n}}=n!\times \sum\limits_{i=0}^{n}{\dfrac{{{\left( -1 \right)}^{n}}}{i!}}$ . We note that in part (iii) we cannot fix the position of P or S on $x=8$ as the position of S or P will be $x+5=8+5=13$ and we do not have 13 positions.