Question

If all the letters of the word ACADEMICIAN are permuted in all possible ways then the number of permutations in which no two A's are together and all the consonants are together is

• A. 7200
• B. 14400
• C. 3600
• D. 1800

Hint:

For "objects not together" problems, use the gap method. First, arrange the objects that do not have restrictions. This creates gaps (including at the ends) where the restricted objects can be placed.

Answer 1

The Correct Option is A

The word is ACADEMICIAN. Total 11 letters.
Vowels: A, A, A, E, I, I (6 total: 3 A's, 2 I's, 1 E).
Consonants: C, D, M, C, N (5 total: 2 C's, 1 D, 1 M, 1 N).
Condition 1: All the consonants are together.
We treat the 5 consonants (C, D, M, C, N) as a single block. The number of ways to arrange the letters within this block is $\frac{5!}{2!} = \frac{120}{2} = 60$ ways.
Condition 2: No two A's are together.
Now we need to arrange this consonant block [CDMC N] along with the remaining vowels which are E, I, I. Let's call the consonant block 'X'. We are arranging X, E, I, I. The number of ways to arrange these 4 items is $\frac{4!}{2!} = \frac{24}{2} = 12$ ways.
This arrangement creates 5 possible gaps where the three A's can be placed so that they are not together. Example arrangement: $_ E _ I _ X _ I _$. We need to choose 3 of these 5 gaps to place the 3 identical A's. The number of ways to do this is $\binom{5}{3} = \frac{5!}{3!2!} = \frac{10}{1} = 10$ ways.
Total number of permutations satisfying both conditions is the product of the number of ways for each step.
Total ways = (Ways to arrange consonants) $\times$ (Ways to arrange the block and other vowels) $\times$ (Ways to place A's).
Total ways = $60 \times 12 \times 10 = 7200$.