Question

If all the letters of the word 'ASSOCIATION' are permuted in all possible ways to form all 11-letter words (with or without meaning), then among these words, the number of words in which the identical letters are always together but no two such groups of identical letters are together is:

• A. \( 1440 \)
• B. \( 2304 \)
• C. \( 576 \)
• D. \( 144 \)

Hint:

To ensure "no two items are together," always arrange the other items first and place the constrained items into the resulting gaps. Since the identical letters within a group like \( \{AA\} \) are the same, there is only \( 1 \) way to arrange letters inside each block.

Answer 1

The Correct Option is D

Concept: This problem involves the "Gap Method" combined with permutations of groups.
• Word: 'ASSOCIATION'. Letters: A(2), S(2), O(2), I(2), C(1), T(1), N(1).
• Treat identical letters as blocks: \( G_1=\{AA\}, G_2=\{SS\}, G_3=\{OO\}, G_4=\{II\} \).
• Singles: \( C, T, N \).
• Constraint: No two \( G_i \) groups can be adjacent.

Step 1: Arranging the single letters.
First, we arrange the 3 single letters \( \{C, T, N\} \). Number of ways = \( 3! = 6 \).

Step 2: Using the Gap Method for groups.
Placing the 3 singles creates 4 gaps: \( \_ \text{C} \_ \text{T} \_ \text{N} \_ \). We must place the 4 blocks \( \{G_1, G_2, G_3, G_4\} \) into these 4 gaps such that no two are together. Number of ways to arrange the 4 blocks in the 4 gaps = \( 4! = 24 \).

Step 3: Total number of ways.
Total arrangements = \( (\text{Ways to arrange singles}) \times (\text{Ways to arrange groups in gaps}) \) \[ \text{Total} = 3! \times 4! = 6 \times 24 = 144 \]