Math Problem Statement

A committee consisting of four faculty members and five students is to be formed. Every committee position has the same duties and voting rights. There are 10 faculty members and 15 students eligible to serve on the committee. In how many ways can the committee be formed?

Solution

To solve this problem, we need to determine the number of ways to form a committee consisting of 4 faculty members and 5 students from a pool of 10 faculty members and 15 students.

Step-by-Step Solution

  1. Determine the number of ways to choose the faculty members:

    We need to choose 4 faculty members from a group of 10. The number of ways to do this is given by the combination formula:

    (104)=10!4!(104)!=10!4!6!\binom{10}{4} = \frac{10!}{4!(10-4)!} = \frac{10!}{4! \cdot 6!}

  2. Determine the number of ways to choose the students:

    Similarly, we need to choose 5 students from a group of 15. The number of ways to do this is also given by the combination formula:

    (155)=15!5!(155)!=15!5!10!\binom{15}{5} = \frac{15!}{5!(15-5)!} = \frac{15!}{5! \cdot 10!}

  3. Calculate the total number of ways to form the committee:

    Since the selections of faculty members and students are independent, the total number of ways to form the committee is the product of the number of ways to choose the faculty members and the number of ways to choose the students:

    Total number of ways=(104)×(155)\text{Total number of ways} = \binom{10}{4} \times \binom{15}{5}

Calculations

Now, let's compute the combinations.

(104)=10×9×8×74×3×2×1=504024=210\binom{10}{4} = \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} = \frac{5040}{24} = 210

(155)=15×14×13×12×115×4×3×2×1=654,120120=5,461\binom{15}{5} = \frac{15 \times 14 \times 13 \times 12 \times 11}{5 \times 4 \times 3 \times 2 \times 1} = \frac{654,120}{120} = 5,461

So, the total number of ways to form the committee is:

210×5,461=1,146,810210 \times 5,461 = 1,146,810

Conclusion

The committee consisting of 4 faculty members and 5 students can be formed in 1,146,810 different ways.

Would you like more details or have any questions?

Here are 5 related questions that might interest you:

  1. How would the number of ways change if we had to select 3 faculty members and 6 students instead?
  2. What is the general formula for combinations, and how is it used in different scenarios?
  3. If we wanted to form multiple committees with the same composition, how would that affect the calculations?
  4. How does the inclusion-exclusion principle apply if some members cannot serve together?
  5. What if we wanted to add constraints, such as at least one faculty member must be from a specific department?

Tip: When dealing with combinations, remember that the order of selection does not matter, unlike permutations where order does matter.