Question

How many different committees can be formed from 99 teachers and 3434 students if the committee consists of 33 teachers and 22 ​students?

Answer 1

There are 47,124 different committees you can select.

Explanation:
The number of combinations of 9 teachers taken 3 at a time is given by
$_9C_3=\frac{9!}{3!6!}=84$

The number of combinations of 34 students taken 2 at a time is given by
$_{34}C_2=\frac{34!}{2!32!}=561$

Together this makes 84*561 = 47124 combinations.

Answer 2

Answer:

47,124 committees.

Step-by-step explanation:

We are asked to find the number of different committees that can be formed from 9 teachers and 34 students if the committee consists of 3 teachers and 2 ​students.

To solve our given problem we will use combination formula:

$_{r}^{n}\textrm{C}=\frac{n!}{r!(n-r)!}$, where,

n= Total number of items,

r = Number of items being chosen at a time.

Since we are choosing 3 teachers from 9 teachers and 2 students from 34 students, so we can represent this information as:

$_{3}^{9}\textrm{C}\times _{2}^{34}\textrm{C}=\frac{9!}{3!(9-3)!}\times \frac{34!}{2!(34-2)!}$

$_{3}^{9}\textrm{C}\times _{2}^{34}\textrm{C}=\frac{9!}{3!(6)!}\times \frac{34!}{2!(32)!}$

$_{3}^{9}\textrm{C}\times _{2}^{34}\textrm{C}=\frac{9*8*7*6!}{3*2*1(6)!}\times \frac{34*33*32!}{2*1(32)!}$

$_{3}^{9}\textrm{C}\times _{2}^{34}\textrm{C}=3*4*7\times 17*33$

$_{3}^{9}\textrm{C}\times _{2}^{34}\textrm{C}=47,124$

Therefore, 47,124 different committees can be formed from 9 teachers and 34 students.