Permutations & Combinations
Arrangements with repetition
Grade 11
Question:
<p>Other than the letter S, the seven letters M, I, I, I, P, P and I can be arranged in \(\dfrac{7!}{2! \cdot 4!} = 7 \cdot 5 \cdot 3\). Now four S can be placed in eight spaces in \({}^8C_4\) ways. Therefore, the required number of ways is \(7 \cdot 5 \cdot 3 \cdot {}^8C_4 = 7 \cdot {}^6C_4 \cdot {}^8C_4\). The required number of ways is:</p>
<p>\(7 \cdot 5 \cdot 3 \cdot {}^8C_4\)</p>
<p>\(7 \cdot {}^6C_4 \cdot {}^8C_4\)</p>
<p>\(7! \cdot {}^8C_4\)</p>
<p>\(7 \cdot 5 \cdot 3 \cdot {}^8C_4\)</p>
Step-by-Step Solution
Key Concept: When arranging letters with repetition and a constraint (like placing S's in specific positions), first arrange the non-constrained letters accounting for their repetitions, then place the constrained letters in the resulting gaps. The 7 letters M,I,I,I,P,P,I create 8 spaces (before first, between each pair, and after last) where S's can be inserted.
<p><strong>Step 1:</strong> Arrange the 7 letters {M, I, I, I, P, P, I} excluding S.</p><p>Number of arrangements = 7!/(4!·2!) = 7·5·3 = 105</p><p><strong>Step 2:</strong> When 7 letters are arranged in a line, they create 8 possible positions for inserting S's: _L₁_L₂_L₃_L₄_L₅_L₆_L₇_</p><p><strong>Step 3:</strong> Choose 4 of these 8 gaps to place the 4 identical S's.</p><p>Number of ways = ⁸C₄ = 70</p><p><strong>Step 4:</strong> Apply multiplication principle (independent choices).</p><p>Total arrangements = (7·5·3) × ⁸C₄ = 105 × 70 = 7,350</p><p><strong>Verification:</strong> 7·5·3·⁸C₄ = 7·⁶C₄·⁸C₄ = 7·15·70 = 7,350</p><p>∴ Answer: D (7,350 or equivalently 7·⁶C₄·⁸C₄)</p>
Correct Answer: D