Questions: How many different committees can be formed from 9 teachers and 46 students if the committee consists of 3 teachers and 3 students?

Solution Steps

To solve this problem, we need to determine the number of ways to form a committee consisting of 3 teachers and 3 students. This involves two separate combinations: choosing 3 teachers from 9 and choosing 3 students from 46. We will calculate the number of combinations for each group and then multiply the results to get the total number of ways to form the committee.

To find the number of ways to choose 3 teachers from a total of 9, we use the combination formula:

\[ \binom{n}{r} = \frac{n!}{r!(n-r)!} \]

For our case, \( n = 9 \) and \( r = 3 \):

\[ \binom{9}{3} = \frac{9!}{3!(9-3)!} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 84 \]

Next, we calculate the number of ways to choose 3 students from a total of 46 using the same combination formula:

\[ \binom{46}{3} = \frac{46!}{3!(46-3)!} = \frac{46 \times 45 \times 44}{3 \times 2 \times 1} = 15180 \]

The total number of ways to form the committee of 3 teachers and 3 students is the product of the two combinations calculated above:

\[ \text{Total Ways} = \binom{9}{3} \times \binom{46}{3} = 84 \times 15180 = 1275120 \]

Final Answer