Question

The number of ways in which a committee of 3 ladies and 4 gentlemen can be appointed from a meeting consisting of 8 ladies and 7 gentlemen, if Mrs. X refuses to serve in a committee if Mr. Y is its member, is

• A. 1960
• B. 3240
• C. 1540
• D. none of these

Hint:

Split into two cases, Mr. Y included or excluded, since Mrs. X's availability depends entirely on whether Mr. Y is on the committee.

Answer 1

The Correct Option is C

To find the number of ways to form a committee of 3 ladies and 4 gentlemen from a group of 8 ladies and 7 gentlemen, while considering the constraint that Mrs. X refuses to serve if Mr. Y is a member, we follow these steps:

1. Identify total selections possible without constraints:
   • Ways to choose 3 ladies out of 8: \(\binom{8}{3}\) 
   • Ways to choose 4 gentlemen out of 7: \(\binom{7}{4}\)
2. Consider the constraint: If Mrs. X is selected, Mr. Y cannot be selected.
   • Calculate the scenario where both Mrs. X and Mr. Y are in the committee:
   • Choose 2 more ladies from the remaining 7: \(\binom{7}{2}\)
   • Choose 3 more gentlemen from the remaining 6: \(\binom{6}{3}\)
3. Subtract the invalid combinations (where both X and Y appear) from total combinations:

Total valid combinations: \(1960 - 420 = 1540\)

Thus, the number of valid ways to form the committee is 1540, which matches the given answer.

Answer 2

Step 1: Split into two cases based on whether Mr. Y is on the committee.
Case 1, Mr. Y is a member: then Mrs. X cannot be on the committee (she refuses whenever Y is a member). So the 3 ladies must be chosen from the remaining 7 ladies (excluding X), and the other 3 gentlemen (besides Y) must be chosen from the remaining 6 gentlemen.
Ways = \( {}^7C_3 \times {}^6C_3 = 35 \times 20 = 700 \)

Step 2: Case 2, Mr. Y is not a member.
Now Mrs. X is free to be included or excluded, so all 8 ladies are available to choose 3 from, and all 4 gentlemen must be chosen from the remaining 6 gentlemen (excluding Y).
Ways = \( {}^8C_3 \times {}^6C_4 = 56 \times 15 = 840 \)

Final Answer:
Total number of ways = 700 + 840 = 1540. The value 1960 is what you'd get by ignoring the X-Y restriction entirely (just \( {}^8C_3 \times {}^7C_4 \)), which is why it's a tempting but wrong answer. \[ \boxed{1540} \]