Question

In how many ways can the letters of the word PERMUTATIONS be arranged so that:
i) Words starting with P and ending with S,
ii) Vowels are all together,
iii) There are always 4 letters between P and S?

Answer 1

Here, we will use the Permutation formula which states that a permutation of any set is known as the arrangement of its members or items into linear order or if that particular set is already ordered rather than a re-arrangement of its elements. For example, we have a set of three numbers, i.e. \left\\{ {1,2,3} \right\\} than there will be six ways to rearrange the elements inside it as shown below:
$\left( {1,2,3} \right)$, $\left( {1,3,2} \right)$, $\left( {2,3,1} \right)$ , $\left( {2,1,3} \right)$ , $\left( {3,2,1} \right)$ and $\left( {3,1,2} \right)$.
So, if there are $n$ numbers of elements inside the set then the permutation formula for re-arranging the elements inside it will be:
${{\text{P}}_n} = n!$ , where ${\text{P}}$ is the permutation symbol, $n$ is the number of items.

Complete step-by-step solution:
Step 1: We need to find the number of ways of rearranging the letters of word PERMUTATIONS, so in the word PERMUTATIONS we will count how many times the letters are repeating:
P(one time), E(one time), R(one time), M(one time), U(one time), T(two times), A(one time), I(one time), O(one time), N(one time), S(one time).
For finding the arrangements in the word PERMUTATIONS when the word starts with P and ends with S, we need to fill the positions of the letters accordingly:
There are total $12$ spaces in the letter PERMUTATIONS, so if the letter is starting by P and ending by S, we have two positions filled i.e. first and last as shown below:

$1^{st}$	$2^{nd}$	$3^{rd}$	$4^{th}$	$5^{th}$	$6^{th}$	$7^{th}$	$8^{th}$	$9^{th}$	$10^{th}$	$11^{th}$	$12^{th}$
P											S

Now, for filling the remaining $10$ positions we will use permutation property, so the arrangement will be:
When the letter starts with P and ends with S= $\dfrac{{10!}}{{2!}}$ ($\because$letter T is repeating twice)
So, the number of arrangements will be =$\dfrac{{10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}}{{2 \times 1}}$ (for example$3! = 3 \times 2 \times 1$)
$\Rightarrow$The number of arrangements (when word starts with the letter P and ends with S) =$1814400$ ways.
Step 2: For finding the arrangements when vowels are together in the word PERMUTATIONS, first we will check how many numbers of vowels are there:
There are $5$ vowels (E, A, U, I, O) in the word PERMUTATIONS and $7$ consonants. Now, for calculating the number of ways first we will consider all the five letters of the vowels as a single letter.
Now except vowels, there are $7$ letters and one is for vowels so there will be total $8$letters to be arranged.
Now we can arrange the $8$ letters as= $8!$
Also, vowels can interchange their positions in five number of ways = $5!$
$\Rightarrow$Total number of ways = $8! \times \dfrac{{5!}}{{2!}}$ (we are dividing by $2!$ because T comes two times in the word)
$\Rightarrow$Total number of ways= $8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 \times \dfrac{{5 \times 4 \times 3 \times 2 \times 1}}{{2 \times 1}}$
By dividing $\dfrac{{5 \times 4 \times 3 \times 2 \times 1}}{{2 \times 1}}$, we get:
$\Rightarrow$Total number of ways= $8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 \times 5 \times 4 \times 3$
By doing simple multiplication of the term $8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 \times 5 \times 4 \times 3$, we get:
$\Rightarrow$Total number of ways (when vowels come together) =$2419200$
Step 3: For finding the number of arrangements in the letter PERMUTATIONS when there will exactly four letters come in between the letters P and S, first we will check that in how many ways we can fill the positions of letters as shown below:
$1^{st}$ way- When the letter P comes at the very first position from starting:
P_ _ _ _ S _ _ _ _ _ _
$2^{nd}$ way- When the letter P comes at 2nd position:
_ P_ _ _ _ S _ _ _ _ _
$3^{rd}$ way- When the letter P comes at 3rd position:
_ _ P _ _ _ _ S _ _ _ _
$4^{th}$ way- When the letter P comes at 4th position:
_ _ _ P _ _ _ _ S _ _ _
$5^{th}$ way: When the letter P comes at 5th position:
_ _ _ _ P _ _ _ _ S _ _
$6^{th}$ way: When the letter P comes at 6th position:
_ _ _ _ _ P _ _ _ _ S _
$7^{th}$ way: When the letter P comes at 7th position:
_ _ _ _ _ _ P _ _ _ _ S
So, there are total $7$ ways for filling the spaces between P and S having $4$ letters between them.
Now, letters P and S can also change their position in the same way. So, again there will be $7$ ways for filling the spaces between S and P having $4$ letters between them.
Hence, P and S or S and P can be filled in $7 + 7 = 14$ many ways.
Also, there are total $12$ spaces which need to be filled and two positions are filled by P and S so remaining positions will be $10$ and the number of ways for filling these positions will be:
$\Rightarrow$Number of ways for filling $10$ positions= $\dfrac{{10!}}{{2!}}$ ( we are dividing by $2!$ because T comes two times in the word)
So, the total number of ways will be equals to the product of the number of ways letter P and S can be filled with the number of ways for filling the remaining $10$ positions:
$\Rightarrow$Total Number of ways (exactly $4$letters are coming in between P and S) = $14 \times \dfrac{{10!}}{{2!}}$
By solving the term $\dfrac{{10!}}{{2!}}$, we get:
$\Rightarrow$Total Number of ways (exactly $4$letters are coming in between P and S) = $14 \times \dfrac{{10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}}{{2 \times 1}}$
By dividing the term $\dfrac{{10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}}{{2 \times 1}}$ we get:
$\Rightarrow$Total Number of ways (exactly $4$letters are coming in between P and S) = $14 \times 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3$
By doing multiplication of the term $14 \times 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3$, we get:
$\Rightarrow$Total Number of ways (exactly $4$ letters are coming in between P and S) = $25401600$

(i) Number of arrangements (when word starts with letter P and ends with S) =$1814400$ ways.
(ii) Total number of ways (when vowels come together) =$2419200$
(iii) Total Number of ways (exactly $4$ letters are coming in between P and S) = $25401600$

Note: In these types of permutation questions students generally get confused in calculating the number of ways for a word when any letter is repeating itself twice or thrice. For example, in a word KAJAL, letter A is repeating itself twice so at the time of finding the number of arrangements of letters we need to divide the result with the repeating letter case as shown below:
In the word KAJAL, there are five letters so the number of ways = $5!$
But letter A is repeating twice so the number of ways will be = $2!$
Total number of ways = $\dfrac{{5!}}{{2!}}$
Also, students need to remember that if we are not selecting any item from the set and taking it out, only re-arrangement is happening then we will use the formula:
Permutation formula= $n!$
But, when we are calculating the number of permutations of $n$the number of objects taken $r$ at a time then we will use the below formula: