Permutations & Combinations
Arrangement with conditions
Grade 11
Question:
<p>Ten IIT and 2 DCE students sit in a row. The number of ways in which exactly 3 IIT students sit between 2 DCE students is</p>
<p>\({}^{10}C_1 \times 2! \times 3! \times 8!\)</p>
<p>\(10! \times 2! \times 3! \times 8!\)</p>
<p>\(5! \times 2! \times 9! \times 8!\)</p>
<p>none of these</p>
Step-by-Step Solution
Key Concept: Treat the 2 DCE students and exactly 3 IIT students between them as a single rigid block, then arrange this block within remaining students while accounting for internal arrangements.
<p><strong>Step 1:</strong> Form a block with 2 DCE students and exactly 3 IIT students between them.</p><p><strong>Step 2:</strong> Choose 3 IIT students from 10 to sit between the 2 DCE students: $\binom{10}{3}$ ways.</p><p><strong>Step 3:</strong> Arrange these 3 chosen IIT students in the middle positions: $3!$ ways.</p><p><strong>Step 4:</strong> Arrange the 2 DCE students on the two ends of the block: $2!$ ways.</p><p><strong>Step 5:</strong> Now we have 1 block + 7 remaining IIT students = 8 entities to arrange in a row: $8!$ ways.</p><p><strong>Step 6:</strong> Total arrangements = $\binom{10}{3} \times 3! \times 2! \times 8!$</p><p>= $120 \times 6 \times 2 \times 40320$</p><p>= $58,060,800$</p><p>∴ Answer: A</p>
Correct Answer: A