To determine how many different committees can be formed from 99 teachers and 3636 students where the committee consists of 33 teachers and 44 students, we need to use the concept of combinations.

Step-by-Step Solution:

1. Combinations Formula: The number of ways to choose $r$ objects from $n$ objects is given by the combination formula: $C(n, r) = \frac{n!}{r!(n - r)!}$

2. Select Teachers: We need to choose 3 teachers from 99 teachers. $C(99, 3) = \frac{99!}{3!(99 - 3)!} = \frac{99!}{3! \cdot 96!}$

3. Select Students: We need to choose 4 students from 3636 students. $C(3636, 4) = \frac{3636!}{4!(3636 - 4)!} = \frac{3636!}{4! \cdot 3632!}$

4. Calculate Each Combination: $C(99, 3) = \frac{99 \times 98 \times 97}{3 \times 2 \times 1} = 156849$ $C(3636, 4) = \frac{3636 \times 3635 \times 3634 \times 3633}{4 \times 3 \times 2 \times 1} = 17366616309$

5. Total Number of Ways to Form the Committee: The total number of different committees is the product of the combinations of teachers and students. $\text{Total ways} = C(99, 3) \times C(3636, 4)$ $\text{Total ways} = 156849 \times 17366616309 = 2722778978105141$

So, the committee of 7 members (3 teachers and 4 students) can be selected in $2,722,778,978,105,141$ different ways.

If you have any further questions or need additional details, feel free to ask!

Suggested Questions:

1. How many ways can you form a committee of 5 people from 12?
2. What is the probability of selecting 2 red cards from a standard deck of cards?
3. How many ways can you arrange the letters in the word "MATHEMATICS"?
4. What is the probability of drawing a face card and a number card from a deck?
5. How many different 3-letter combinations can be formed from the alphabet?

Tip:

When dealing with combinations, always ensure to distinguish between combination and permutation problems. Combinations are used when the order does not matter, while permutations are used when the order does matter.