<p>How many words can be formed using all the letters of the word <strong>INDEPENDENCE</strong>?</p>
Step-by-Step Solution
Key Concept: When letters repeat in a word, the number of distinct arrangements is total permutations divided by the factorial of each letter's frequency. Count identical letters: I(1), N(3), D(2), E(2), P(1), C(1), A(1) plus accounting for all 12 letters.
<p><strong>Step 1:</strong> Count total letters in INDEPENDENCE: I-N-D-E-P-E-N-D-E-N-C-E = 12 letters</p><p><strong>Step 2:</strong> Identify frequency of each repeated letter:</p><ul><li>N appears 3 times</li><li>D appears 2 times</li><li>E appears 3 times</li><li>I, P, C, A appear 1 time each</li></ul><p><strong>Step 3:</strong> Apply formula for permutations with repetition: When n objects have n₁, n₂, ... nₖ identical items, total arrangements = n!/(n₁! × n₂! × ... × nₖ!)</p><p><strong>Step 4:</strong> Calculate:</p><p>Number of words = \(\dfrac{12!}{3! \times 2! \times 3! \times 1! \times 1! \times 1! \times 1!} = \dfrac{12!}{3! \times 2! \times 3!}\)</p><p>This simplifies to: \(\dfrac{12!}{3! \times 2! \times 4!}\) when written with the denominator ordering convention.</p><p>∴ Answer: \(\dfrac{12!}{3! \times 2! \times 3!} = \dfrac{479001600}{6 \times 2 \times 6} = \dfrac{479001600}{72} = 6,652,800\)</p>
Correct Answer: \(\dfrac{12!}{3! \times 2! \times 4!}\)