Question

Number of different words that can be formed using all the letters of the word "DEEPMALA" if two vowels are together and the other two are also together but separated from the first two is

• A. 960
• B. 1200
• C. 2160
• D. 1440

Answer 1

Answer: (D)

1440

Answer 2

Here's how to solve this permutations problem with restrictions:

1. Identify Letters:

   The word DEEPMALA has 8 letters: 4 consonants (D, P, M, L) and 4 vowels (E, E, A, A).

2. Form Vowel Blocks:

   The condition requires forming two blocks of two vowels each. There are two distinct ways to group the vowels:

   • Case 1: Blocks are (EE) and (AA).

     • The block (EE) can be formed in 1 way (since Es are identical).
     • The block (AA) can be formed in 1 way (since As are identical).
     • These two blocks, (EE) and (AA), are distinct.

   • Case 2: Blocks are (EA) and (EA).

     • The first block, (EA), can be formed by choosing one E and one A. Its internal arrangement can be EA or AE (2 ways).
     • The second block, (EA), can also be formed in 2 ways (EA or AE).
     • So, there are $2 \times 2 = 4$ ways to form the specific strings for these two blocks (e.g., (EA)(EA), (EA)(AE), (AE)(EA), (AE)(AE)).
     • For the purpose of placement in the overall word, if the two blocks have the same composition (both are {E,A}), they are considered identical units.

3. Arrange Consonants and Vowel Blocks:

   • Arrange the 4 distinct consonants (D, P, M, L): $4! = 24$ ways.

   • These 4 consonants create 5 gaps where the vowel blocks can be placed: _ C _ C _ C _ C _.

   • The two vowel blocks must be placed in different gaps (separated).

   • For Case 1 (Blocks (EE) and (AA)):

     • Since (EE) and (AA) are distinct blocks, they can be placed in 2 chosen gaps in $\binom{5}{2} \times 2!$ ways.
     • Number of ways to choose 2 gaps from 5 = $\binom{5}{2} = 10$.
     • Number of ways to arrange the 2 distinct blocks in these 2 gaps = $2! = 2$.
     • Total placement ways = $10 \times 2 = 20$.
     • Total words for Case 1 = (Arrangement of consonants) $\times$ (Placement of blocks) $\times$ (Internal arrangements) $= 24 \times 20 \times (1 \times 1) = 480$.

   • For Case 2 (Blocks (EA) and (EA)):

     • The internal arrangements for the two (EA) blocks give $2 \times 2 = 4$ distinct string pairs (e.g., (EA)(EA), (EA)(AE)).
     • However, since the composition of the two blocks is the same ({E,A} and {E,A}), they are treated as identical units for placement in the gaps.
     • Number of ways to choose 2 gaps from 5 for these identical blocks = $\binom{5}{2} = 10$. (No $2!$ because swapping identical blocks doesn't create a new arrangement).
     • Total placement ways = $10$.
     • Total words for Case 2 = (Arrangement of consonants) $\times$ (Placement of blocks) $\times$ (Internal arrangements) $= 24 \times 10 \times (2 \times 2) = 24 \times 10 \times 4 = 960$.

4. Total Number of Words:

   Sum the results from Case 1 and Case 2:

   Total words = $480 + 960 = 1440$.

Therefore, the final answer is 1440.