Question

in how many ways can a committee of 5 be formed from agroup of 11 people consisting of 4 teachers and 7 students if
- there is no restriction in the selection?
- the committee must include exactly 2 teachers?

Answer 1

Answer on question 34134 – Math – Combinatorics

in how many ways can a committee of 5 be formed from agroup of 11 people consisting of 4 teachers and 7 students if

- there is no restriction in the selection?

- the committee must include exactly 2 teachers?

Solution

We have 11 people. The number of ways to choose 5 of them is

$$C_{11}^{5} = \frac{11!}{5!(11-5)!} = \frac{11 \times 10 \times 9 \times 8 \times 7}{2 \times 3 \times 4 \times 5} = 462.$$

the number of choosing 2 teachers from 4 is

$$C_{4}^{2} = \frac{4!}{2! \cdot 2!} = 6$$

And the number of ways to choose 3 students of 7 is

$$C_{7}^{3} = \frac{7!}{3! \cdot 4!} = \frac{7 \times 6 \times 5}{6} = 35.$$

If committee must include exactly 2 teachers then the number of ways to choose such committee is 6*35=210.

Answer: 462 and 210.

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