Forming a Committee of 5: Constraints and Calculations
In many scenarios, forming committees requires adherence to specific constraints. This article explores a common problem: forming a committee of 5 from a group of 11 people consisting of 4 teachers and 7 students, with a specific constraint: a certain teacher and student cannot both be in the committee. We will walk through the steps to solve this problem using combinatorics.Step-by-Step Solution
To solve this problem, we need to calculate the total number of ways to form a committee of 5 people from a group of 11 people (4 teachers and 7 students) without any restrictions, and then subtract the number of unrestricted cases where both the specific teacher and student are included. Total Ways Without Restrictions First, we calculate the total number of ways to form a committee of 5 from 11 people using the combination formula:binom{n}{r} frac{n!}{r!(n-r)!}
Here, n 11 and r 5.Therefore, binom{11}{5} frac{11!}{5!6!} frac{11 times 10 times 9 times 8}{4 times 3 times 2 times 1} 462.
Restrictions and Calculation
Next, we need to subtract the cases where both the specific teacher (let's call them Teacher A) and the specific student (Student B) are included in the committee. If both are included, we then need to choose 3 more members from the remaining 9 people (3 teachers and 6 students). This can be calculated as:binom{9}{3} frac{9!}{3!6!} frac{9 times 8 times 7}{3 times 2 times 1} 84.
Final Calculation Finally, we subtract the restricted cases from the total cases to get the number of valid committees:text{Total valid committees} text{Total committees} - text{Committees with Teacher A and Student B}
text{Total valid committees} 462 - 84 378.
Conclusion
Thus, the number of ways to form a committee of 5 people from the group of 11, ensuring that Teacher A and Student B are not both included, is 378.For a simpler calculation:
Total possible 5-person committees with 11 people: binom{11}{5} 462. Committees with the indicated teacher and student on the committee: binom{9-2}{5-2} binom{9}{3} 84. Required number of ways: 462 - 84 378.