Permutations & Combinations
Selection with Multiple Conditions
Grade 11
Question:
<p>A man X has 7 friends, 4 of them are ladies and 3 are men. His wife Y also has 7 friends, 3 of them are ladies and 4 are men. Assume X and Y have no common friends. Then the total number of ways in which X and Y together can throw a party inviting 3 ladies and 3 men, so that 3 friends of each of X and Y are in this party, is</p>
<p>(a) 484</p>
<p>(b) 485</p>
<p>(c) 468</p>
<p>(d) 469</p>
Step-by-Step Solution
Key Concept: Partition invitations into cases based on how many ladies and men each person invites, ensuring the total is 3 ladies and 3 men with each person inviting exactly 3 friends.
<p><strong>Solution:</strong> Total guests invited: 3 ladies and 3 men, with 3 friends from X and 3 from Y.</p><p>X's friends: 4 ladies, 3 men</p><p>Y's friends: 3 ladies, 4 men</p><p><strong>Case 1:</strong> X invites 2 ladies, 1 man; Y invites 1 lady, 2 men</p><p>Ways: $\binom{4}{2} \times \binom{3}{1} \times \binom{3}{1} \times \binom{4}{2} = 6 \times 3 \times 3 \times 6 = 324$</p><p><strong>Case 2:</strong> X invites 1 lady, 2 men; Y invites 2 ladies, 1 man</p><p>Ways: $\binom{4}{1} \times \binom{3}{2} \times \binom{3}{2} \times \binom{4}{1} = 4 \times 3 \times 3 \times 4 = 144$</p><p><strong>Case 3:</strong> X invites 3 ladies, 0 men; Y invites 0 ladies, 3 men</p><p>Ways: $\binom{4}{3} \times \binom{3}{0} \times \binom{3}{0} \times \binom{4}{3} = 4 \times 1 \times 1 \times 4 = 16$</p><p><strong>Case 4:</strong> X invites 0 ladies, 3 men; Y invites 3 ladies, 0 men</p><p>Ways: $\binom{4}{0} \times \binom{3}{3} \times \binom{3}{3} \times \binom{4}{0} = 1 \times 1 \times 1 \times 1 = 0$ (Not possible)</p><p>Total: $324 + 144 + 16 = 484$</p><p>∴ Answer is (a) 484.</p>
Correct Answer: A