Question

How many different words can be formed from the letters of the word INTERMEDIATE? In how many of them, two vowels never come together?

Answer 1

Hint: Make use of permutations to arrange the letters (12 in this case). Use the multiplication of factorials to incorporate the repetition of letters. Use the property of permutation, n places can be filled by r distinct items in ${}^{n}{{\operatorname{P}}_{r}}$ ways, that is, 7 places can be filled by 6 distinct items in ${}^{7}{{\operatorname{P}}_{6}}$ ways. To find the permutations in which the vowels never come together, we fix the six vowels with gaps between them which are to be filled by the six consonants

Complete step-by-step solution:
If the 12 letters of the given word were distinct, we could have easily found out the answer using multiplication theory. The answer would have been 12!. Since there are multiple occurrences of different letters (indistinct) we use a different equation. We find the product of factorials of the number of occurrences of the different letters in the word and then divide the total number of possibilities (12! in this particular case ) by this product.
Let us first consider part one of the question. Here, we are asked to find the different words (with or without meaning and having 12 letters) that can be formed from the letters of the word INTERMEDIATE.
Total number of words that can be formed = (number of letters)! = 12!
Now if we look at the different letters of the word INTERMEDIATE, we can find that it has only 8 different letters namely I, N, T, E, R, M, D, and A. We can also see that the letter E is used thrice, I and T have used twice each and the rest of the letters are used once each.
By using multiplication theory,
 Product representing repetition = (number of E)! $\times $(number of I)! $\times $(number of T)!
=2!2!3! (The rest of the letters’ occurrences contribute 1! each to the product )
$\therefore $ The number of different words that can be formed from the letters of the word INTERMEDIATE$=\dfrac{12!}{2!2!3!}=12\times 11\times 10\times 9\times 8\times 7\times 6\times 5=19958400$
Hence the total number of words that can be formed using the letters of the word INTERMEDIATE is $=\dfrac{12!}{2!2!3!}=19958400$
Let us now consider part two of the question.
Here we are asked to find the number of words in which the vowels do not come together. We have to find the number of ways the consonants can be arranged and then multiply it with the number of ways the vowels can be arranged. This gives us the total number of ways the letters can be arranged when the vowels do not come together.
Number of ways consonants can be arranged $=\dfrac{6!}{2!1!1!1!1!}=360$
Fixing 6 consonants creates 7 gaps which are to be filled by the 6 vowels (3 E, 2 I, and 1 A)
If all the vowels were different the 7 gaps can be arranged in ${}^{7}{{P}_{6}}$ ways.
But, since E and I occur thrice and twice respectively,
Number of ways vowels can be arranged =$\dfrac{{}^{7}{{P}_{6}}}{2!3!}$ $=\dfrac{7!}{1!2!3!}=\dfrac{5040}{12}=420$
$\therefore $ Total number of ways the letters can be arranged such that two vowels never come together
 = 420 $\times $ 360 = 151200
Hence, the total number of ways in which the letters of the word INTERMEDIATE can be arranged such that no vowels come together = 151200

Note: Always be careful while doing calculations. Never miss out on multiple occurrences of letters.
Alternate method:
Filling of 7 gaps by the 6 vowels can also be seen as an arrangement of a set of 6 into 7 slots, that is 7! ways are possible (Here the use of permutation can be avoided).