The number of different words that can be made from the letters of the word INTERMEDIATE,

Solution:

Number of letters in INTERMEDIATE is 12. 

Number of vowels $(A, E, E, I, I, E)$ i.e is 6 .

Number of consonants (T T . R M N D) i.e is $6 .$

$\therefore$ Total words are $\frac{12 !}{3 ! 2 ! 2 !}$

Now, number of ways of arranging 6 consonants (2 alike) is

$\frac{6 !}{2 !}=6 \times 5 \times 4 \times 3$

= 360

There are 7 gaps in which 6 vowels can be arranged in 7P6 ways but 2 are alike of are kind and 3 of other kind

$\therefore$ Number of ways of arranging the vowels is ${ }^{7} P_{6} \times \frac{1}{3 ! 21}$

$=\frac{7 !}{1 !} \times \frac{1}{3 ! 2 !}$

$=\frac{7 !}{3 \times 2 \times 2}=\frac{7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{3 \times 2 \times 2}$

$=20 \times 21$

$=420$

Hence, the total number of ways when the two vowels never come together is

360 × 420

= 151200