Permutations & Combinations
Selection with Conditions
Grade None
Question:
<p>A group of students comprises of 5 boys and <em>n</em> girls. If the number of ways, in which a team of 3 students can randomly be selected from this group such that there is at least one boy and at least one girl in each team, is 1750, then <em>n</em> is equal to: <strong>(JEE Main 2019, April)</strong></p>
<p>28</p>
<p>27</p>
<p>25</p>
<p>24</p>
Step-by-Step Solution
Key Concept: The constraint 'at least one boy and at least one girl' means we select teams of type (2B,1G) or (1B,2G). Set up the equation C(5,2)×C(n,1) + C(5,1)×C(n,2) = 1750 and solve the resulting quadratic.
<p><strong>Step 1:</strong> Identify valid team compositions with at least 1 boy and 1 girl from 3 students:</p><ul><li>Case 1: 2 boys and 1 girl</li><li>Case 2: 1 boy and 2 girls</li></ul><p><strong>Step 2:</strong> Write the equation for total favorable outcomes:</p><p>C(5,2) × C(n,1) + C(5,1) × C(n,2) = 1750</p><p><strong>Step 3:</strong> Calculate combinations:</p><p>10 × n + 5 × (n(n-1)/2) = 1750</p><p>10n + (5n² - 5n)/2 = 1750</p><p><strong>Step 4:</strong> Simplify by multiplying by 2:</p><p>20n + 5n² - 5n = 3500</p><p>5n² + 15n = 3500</p><p>n² + 3n = 700</p><p>n² + 3n - 700 = 0</p><p><strong>Step 5:</strong> Solve using the quadratic formula or factoring:</p><p>(n + 28)(n - 25) = 0</p><p>n = 25 (taking positive value)</p><p><strong>Verification:</strong> 10(25) + 5(25)(24)/2 = 250 + 1500 = 1750 ✓</p><p>∴ Answer: <strong>n = 25</strong></p>
Correct Answer: C