<p>Letters of the word INTERNATIONAL are (I, I), (N, N, N), (T, T), (A, A), E, R, O, L. Which of the following are correct about the number of ways to select 5 different letters?</p>
<p>(a) 5 distinct letters can be chosen in \({}^8C_5 = 56\) ways</p>
<p>(b) Two identical letters and 3 distinct letters can be chosen in \(({}^4C_1)({}^7C_3) = 140\) ways</p>
<p>(c) Two sets of identical letters and one distinct letter can be chosen in \(({}^4C_2)({}^6C_1) = 36\) ways (actually it equals 36)</p>
<p>(d) Total number of required ways is 242</p>
Step-by-Step Solution
Key Concept: When selecting 5 different letters from a multiset, we must first choose which 5 distinct letters from the available letter types {I, N, T, A, E, R, O, L}, then account for how many ways we can pick them considering multiplicities. The constraint 'different letters' means each selected letter must be distinct in identity, not that we pick one from each multiplicity.
<p><strong>Step 1: Identify available letter types</strong></p><p>Letter types with their frequencies: I(2), N(3), T(2), A(2), E(1), R(1), O(1), L(1)</p><p>Total of 8 distinct letter types available.</p><p><strong>Step 2: Understand 'select 5 different letters'</strong></p><p>We must choose 5 distinct letter types from these 8 types: C(8,5) = 56 combinations.</p><p><strong>Step 3: Analyze specific combinations</strong></p><p>Once we select which 5 letter types, the number of ways to actually pick them depends on their multiplicities:</p><p>• If all 5 selected letters have multiplicity 1: only 1 way to select (e.g., {E,R,O,L,I} → 1 way)</p><p>• If k letters have multiplicity ≥2: we have 2^k ways (choose which instance of each repeated letter)</p><p><strong>Step 4: Categorize selections</strong></p><p>• Selections including only single-occurrence letters {E,R,O,L}: C(4,5) = 0 (impossible)</p><p>• Selections from {I,N,T,A,E,R,O,L} with exactly 1 letter from {I,N,T,A}: C(4,1)×C(4,4) = 4 ways with 2¹ = 2 selections each = 8 total</p><p>• Selections with 2 letters from {I,N,T,A}: C(4,2)×C(4,3) = 6×4 = 24 ways with 2² = 4 selections each = 96 total</p><p>• Selections with 3+ letters from {I,N,T,A}: generate further multiplicities</p><p><strong>Total ways = 56 ways if counting just letter type combinations, or 2^k adjustment for multiplicities within selections</strong></p><p>∴ Answer: A, C</p>
Correct Answer: A, C