<p>If 'K' is the number of ways in which we can choose 5 letters from the word INTERNATIONAL then-</p>
<p>(a) K is a three digit number</p>
<p>(b) K is divisible by 4</p>
<p>(c) K is divisible by 11</p>
<p>(d) k is divisible by 9</p>
Step-by-Step Solution
Key Concept: The word INTERNATIONAL contains repeated letters (I appears 2 times, N appears 2 times), so we must use the stars-and-bars method with constraints to count distinct 5-letter selections, not simple combinations.
<p><strong>Step 1: Identify the available letters in INTERNATIONAL</strong></p><p>Letters: I(2), N(2), T(1), E(1), R(1), A(1), O(1), L(1) — 11 distinct positions, 8 distinct letters</p><p><strong>Step 2: Set up cases based on repetition patterns for selecting 5 letters</strong></p><p>We need 5 letters from {I, N, T, E, R, A, O, L} where I and N can appear 0, 1, or 2 times; others appear 0 or 1 time.</p><p><strong>Step 3: Enumerate by cases of repeated letters</strong></p><p><strong>Case 1:</strong> Both I and N appear twice: I²N²(one from 6 others) = 6 ways</p><p><strong>Case 2:</strong> I appears twice, N once: I²N(3 from 6 others) = C(6,3) = 20 ways</p><p><strong>Case 3:</strong> N appears twice, I once: N²I(3 from 6 others) = C(6,3) = 20 ways</p><p><strong>Case 4:</strong> I appears twice, N doesn't: I²(3 from 6 others) = C(6,3) = 20 ways</p><p><strong>Case 5:</strong> N appears twice, I doesn't: N²(3 from 6 others) = C(6,3) = 20 ways</p><p><strong>Case 6:</strong> I appears once, N appears once: IN(3 from 6 others) = C(6,3) = 20 ways</p><p><strong>Case 7:</strong> Neither I nor N appears: C(6,5) = 6 ways</p><p><strong>Step 4: Add all cases</strong></p><p>K = 6 + 20 + 20 + 20 + 20 + 20 + 6 = <strong>112</strong></p><p>∴ Answer: A</p>
Correct Answer: A