Question

In how many ways can the letters of the word PERMUTATIONS be arranged if the
1) Words start with P and ends with s and
2) Vowels are all together.

Answer 1

Hint:
We use the permutation formula ${p_n} = n!$ where p is the permutation symbol and n is the number of items. A permutation is an arrangement of items into linear order.

Complete Step by Step Solution:
The objective of the problem is to find in how ways can the letter of the PERMUTATIONS be arranged if words starts with P and ends with S and vowels all are together
For the first one that words start with P and end with S we need to find the number of ways of rearranging the letters in the word PERMUTATIONS. In the given word T is the only letter repeating two times and remaining letters are repeating only once .
For the first question , there are 12 questions in the given letter if the letter is starting with P and ending with S we have two positions to be filled. So there are 10 places to be filled .
Therefore the required permutation is $\dfrac{{10!}}{{2!}}$ we wrote 2! because T is repeated two times. The number of arrangements if the words start with P and ends with S will be $\dfrac{{10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}}{{2 \times 1}} = 181440$
For the second one that is to find in how many ways can the vowels all together of the word PERMUTATIONS be arranged. For this first we should count the number of vowels in the given word PERMUTATIONS. There are five vowels in the given word. First consider all the five vowels as one letter and from the remaining 7 letters and one letter is for vowels . so totally there are 8 letters.
The eight letters can be arranged in 8!. As there are five vowels , these five vowels can interchange their positions in 5! Ways.
Therefore the total number of ways that all vowels can be together is $8! \times \dfrac{{5!}}{{2!}}$
$
   = 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}} \\
   = 2419200 \\
 $

Thus , Number of arrangements if the words start with P and ends with S is 181440.
Number of ways that all vowels can be together is 2419200.

Note:
In this type of repeating letters questions we should divide the result with the repeating number of letters. There can be many variations of the same question. A minor change in question will lead to a different question.