Question

The number of words that can be formed from the letters of the word "INTERMEDIATE" in which the vowels are never together is:

• A. $6!{}^7{P}_6$
• B. $\frac{6!}{2!}\times \frac{{}^7{P}_6}{2!3!}$
• C. $\frac{6!\times {}^7{P}_6}{2!3!}$
• D. $\frac{(7!)\times {}^7{P}_6}{2!5!}$

Answer 1

The correct option is B $\frac{6!}{2!}\times \frac{{}^7{P}_6}{2!3!}$

INTERMEDIATE

I $\to$ 2 Consonants - 6, T-2,N-1,R-1,M-1,D-1
N $\to$ 1 Vowels - 6, I-2, A-1, E-3
T $\to$ 2
E $\to$ 3
R $\to$ 1
M $\to$ 1
D $\to$ 1
A $\to$ 1
Now, for rearrangements.
$C_1$ x $C_2$ x $C_3$ x $C_4$ x $C_5$ x $C_6$
$\to$ Now, the Vowels could appear in any six of these 7 places. So, first we choose 6 out of 7 places = ${}^7{C}_6$
$\to$ Now, no. of ways of rearranging these 6 vowels is = $\frac{6!}{2!1!3!}=\frac{6!}{2!3!}$
$\to$ No. of ways of rearranging the consonants, is $\frac{6!}{2!1!1!1!1!}=\frac{6!}{2!}$
Total no. of ways = $\frac{6!}{2!}\times \frac{6!}{2!2!}{\times }^7{C}_6$
$=\frac{6!}{2!}\times \frac{{}^7{P}_6}{2!3!}$