Question

Find the number of words formed by permuting all letters of the word INDIA. How many of them have vowels together?

Answer 1

Hint: First calculate the number of cases for which 5 different letters can be arranged. Now add the condition that 2 of them are identical. By this we will get a number of ways of arranging the words. Now combine the vowels and make them as one set. Now by treating the whole set as one letter arrange them all. After that, apply the condition of repeating vowels to get the required result.

Complete step-by-step answer:
Given word in the question can be written as follows:
I N D I A
For now let us assume the I’s are different, we get:
${I_1}\,N\,\,D\,\,{I_2}\,A$
Factorial: In mathematics, factorial is an operation, denoted by “(!)”. It represents the product of all numbers between 1 and a given number.
In simple words factorial of a number can be found by multiplying all terms you get by subtracting 1 from a given number repeatedly till you get the number difference as 1. Its representation can be written as:
$n! = n \times \left( {n - 1} \right) \times \left( {n - 2} \right)............. \times 1$
For example: $5! = 5 \times 4 \times 3 \times 2 = 120$
Note that we assume $0! = 1$. It is standard value. It has a wide range of applications in combinations.
For arranging the above number, use basic property of arranging. To arrange n particles, we calculate ways as:
Number of possibilities for ${1^{st}}$ particle = n places
Number of possibilities after 1 place filled = $\left( {n - 1} \right)$ places
,
,
,
Number of possibilities for last particle = 1 place.
By product rule we get number of ways given by:
Total ways $= \left( n \right) \times \left( {n - 1} \right)........ \times 1$
By definition of factorial we can say the above equation as:
Total ways $= 5!$
So, number of ways of arranging ${I_1}\,N\,\,D\,\,{I_2}\,A$ are given by:
$5!$
For every arrangement if ${I_1} = {I_2} = I$ we are counting each arrangement twice. So, total numbers of ways are given by:
$\dfrac{{5!}}{{2!}}\,ways\, = 5 \times 4 \times 3$
By simplifying, we get number of words possible as follows:
Total number of words$= {\rm{6}}0$.
Now by taking all vowels together, we write letters as:
${I_1},\,{I_2},\,A,N,\,\,D\,\,$
By combining the first three letters, we get it as:
${I_1}\,{I_2}\,A,N,\,\,D\,\,$
Total ways to arrange 3 letters can be written as:
Ways 1 $= 3!$
Now inside combined vowels we have 3 letters. So it is:
$3!$
As 2 I’s are same, each one is counted twice, we get:
Ways 2 $= \dfrac{{3!}}{{2!}}$
By applying product rule we get total ways $= 3!\, \times \dfrac{{3!}}{{2!}}$
By simplifying, we get total words of vowel together$= \dfrac{{36}}{2} = 18$
Therefore total numbers of words possible are 60 words out of which 18 words have vowels together.

Note: Don’t forget to consider the repetitions. Even when the vowels are combined don’t forget the arrangements of vowels inside the combined part, inside that also you must consider repetition. The idea of assuming repeated terms as different and again making them identical is very important which helps for better understanding.