How Many Ways Can a Committee of 5 Teachers and 4 Students Be Chosen from 9 Teachers and 15 Students? A Comprehensive Guide

Introduction

Determining the number of ways to form a specific committee from a larger group is a common problem in combinatorics. In this article, we will explore the process of selecting a committee of 5 teachers and 4 students from 9 teachers and 15 students using the principle of combinations and binomial coefficients. We will also discuss variations of this problem and provide step-by-step solutions for clarity and understanding.

Combination Basics

Before diving into the problem, it is essential to understand the concept of combinations. A combination is a selection of items where the order does not matter. The formula for combinations, often denoted as ( _n C_r ) or ( binom{n}{r} ), is given by:

$$ _n C_r binom{n}{r} frac{n!}{r!(n-r)!} $$

Where ( n! ) (n factorial) represents the product of all positive integers up to ( n ).

Selecting Teachers

To form a committee with 5 teachers from 9, we use the combination formula with ( n 9 ) and ( r 5 ):

$$ _9 C_5 binom{9}{5} frac{9!}{5!(9-5)!} frac{9!}{5!4!} $$

Calculating this, we get:

$$ binom{9}{5} frac{9 times 8 times 7 times 6}{4 times 3 times 2 times 1} 126 $$

Selecting Students

Similarly, to choose 4 students from 15, we use the combination formula with ( n 15 ) and ( r 4 ):

$$ _{15} C_4 binom{15}{4} frac{15!}{4!(15-4)!} frac{15!}{4!11!} $$

Calculating this, we get:

$$ binom{15}{4} frac{15 times 14 times 13 times 12}{4 times 3 times 2 times 1} 1365 $$

Total Combinations

The total number of ways to form the committee is the product of the two combinations:

$$ text{Total ways} binom{9}{5} times binom{15}{4} 126 times 1365 $$

Calculating this product, we get:

$$ 126 times 1365 171990 $$

Thus, the total number of ways to choose a committee of 5 teachers and 4 students from 9 teachers and 15 students is 171,990.

Variations and Further Explorations

Let's explore variations of this problem. For instance, if student B is selected and teacher A refuses to participate:

Student B Selected, Teacher A Refused

First, we calculate the total number of committees without any restrictions:

$$ _9 C_5 times _{15} C_4 171990 $$

Next, we account for the scenario where both B and A are included in the committee:

Choosing 4 teachers from the remaining 8 teachers and 3 students from the remaining 14 students:

$$ _8 C_4 times _{14} C_3 25480 $$

The difference provides the number of valid committees:

$$ 171990 - 25480 146510 $$

Conclusion

In conclusion, the number of ways to form a committee of 5 teachers and 4 students from 9 teachers and 15 students is 171,990. This problem demonstrates the application of combinations and binomial coefficients, providing a fundamental understanding of selecting groups from larger sets.