<p>A team of four students is to be selected from a total of 12 students. The total number of ways in which the team can be selected such that two particular students refuse to be together and other two particular students wish to be together only is equal to</p>
Step-by-Step Solution
Key Concept: Use complementary counting and conditional selection: subtract cases where the two refusing students are together, then ensure the two willing students are always together as a single unit.
<p><strong>Step 1:</strong> Identify the constraints:</p><ul><li>Two students (say A, B) refuse to be together</li><li>Two students (say C, D) wish to be together only (must both be in team or both out)</li></ul><p><strong>Step 2:</strong> Total ways without any restriction = C(12,4) = 495</p><p><strong>Step 3:</strong> Apply condition on C and D (must be together or not at all):</p><ul><li>Case I: Both C and D are in the team → Select C, D, and 2 from remaining 10 = C(10,2) = 45</li><li>Case II: Neither C nor D in the team → Select 4 from remaining 10 = C(10,4) = 210</li><li>Valid selections with C,D condition = 45 + 210 = 255</li></ul><p><strong>Step 4:</strong> From these 255, subtract cases where both A and B are together:</p><ul><li>Cases with both A, B, C, D: = 1 way</li><li>Cases with A, B only (C, D both out) → Select A, B and 2 from remaining 8 = C(8,2) = 28</li><li>Total invalid = 1 + 28 = 29</li></ul><p><strong>Step 5:</strong> Required answer = 255 - 29 = <strong>226</strong></p><p>∴ Answer: C</p>
Correct Answer: C