Question

Find the number of words formed by permuting all the letters of the word INDEPENDENCE such that the E’s do not come together.

• A. 24300
• B. 1632960
• C. 1663200
• D. 30240
• E. 12530

Hint:

For “not together” problems, use: Total arrangements – Arrangements (together).

Answer 1

The Correct Option is B

Step 1: Total arrangements without restriction.
Word: INDEPENDENCE (12 letters).
Letter frequencies: I – 1, N – 3, D – 2, E – 4, P – 1, C – 1.
Total arrangements: \[ \frac{12!}{3! \cdot 2! \cdot 4!} = \frac{479001600}{288} = 1663200 \]
Step 2: Count arrangements with all 4 E’s together.
Treat 4 E’s as one block. Then total symbols = \(12 - 4 + 1 = 9\).
Frequencies now: N – 3, D – 2, (E-block) – 1, I – 1, P – 1, C – 1.
Total = \[ \frac{9!}{3! \cdot 2!} = \frac{362880}{12} = 30240 \]
Step 3: Subtract.
Words with E’s not together = \(1663200 - 30240 = 1632960\).
Final Answer:
\[ \boxed{1632960} \]