<p>Let \(S = \{1, 2, 3, \ldots, 9\}\). For \(k = 1, 2, \ldots, 5\), let \(N_k\) be the number of subsets of <i>S</i>, each containing five elements out of which exactly <i>k</i> are odd. Then \(N_1 + N_2 + N_3 + N_4 + N_5 =\)</p>
Step-by-Step Solution
Key Concept: S has 5 odd numbers {1,3,5,7,9} and 4 even numbers {2,4,6,8}. For a 5-element subset with exactly k odd numbers, we need k odd and (5-k) even numbers, so we calculate C(5,k)·C(4,5-k) for each valid k and sum them.
<p><strong>Step 1:</strong> Identify the composition of S. S has 5 odd numbers {1,3,5,7,9} and 4 even numbers {2,4,6,8}.</p><p><strong>Step 2:</strong> For a 5-element subset with exactly k odd numbers, we need exactly k odd and (5-k) even numbers.</p><p><strong>Step 3:</strong> Calculate each N_k:</p><ul><li>N₁ = C(5,1)·C(4,4) = 5·1 = 5</li><li>N₂ = C(5,2)·C(4,3) = 10·4 = 40</li><li>N₃ = C(5,3)·C(4,2) = 10·6 = 60</li><li>N₄ = C(5,4)·C(4,1) = 5·4 = 20</li><li>N₅ = C(5,5)·C(4,0) = 1·1 = 1</li></ul><p><strong>Step 4:</strong> Sum all values: N₁ + N₂ + N₃ + N₄ + N₅ = 5 + 40 + 60 + 20 + 1 = 126</p><p><strong>Verification:</strong> This equals C(9,5) = 126, which is the total number of 5-element subsets of S (confirming our partition is exhaustive and mutually exclusive).</p><p>∴ Answer: <strong>126</strong></p>
Correct Answer: B