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

No. of letters $=12;6$ vowels $(2l,3E,1A)$ and 6 consonants $(2T,1R,1M,1N,1D)$.
Total words $=\frac{12!}{2!3!2!}$
Gap Method:
No. of ways of arranging 6 consonants (2 alike) is
$\frac{6!}{2}=360$ ....(1)
There will be 7 gaps in which 6 vowels if all different can be arranged in ${}^7{P}_6$ ways but a 2 are alike of one kind and 3 of other kind.
$\therefore$ Number of arranging the vowels is
${}^7{P}_6\cdot \frac{1}{3!2!}=\frac{7!}{12}=\frac{5040}{12}=420$ ....(2)
Hence the total number of ways when the two vowels never come together by fundamental theorem is $360\times 420=151200$ by (1), (2)