How many different committees can be formed from 10 teachers and 37 students of the committee consist of three teachers and two students? how many ways can the committee of five members be selected?

79,920 different ways

Step-by-step explanation:

Combination has to do with selection:

If we are to select 3 teachers from a pool of 10 teachers to form a committee, this can be done in 10C3 number of ways.

10C3 = 10!/(10-3)!3!

10C3 = 10!/7!3!

10C3 = 10*9*8*7!/7!3!

10C3 = 10*9*8/3*2

10C3 = 720/6

10C3 = 120 ways

Similarly, selecting five 2 students from a pool of 37 students to form a committee, this can be done in 37C2 difference ways;

37C2 = 37!/(37-2)!2!

37C2 = 37!/35!2!

37C2 = 37*36*35!/35!*2

37C2 = 37*36/2

37C2 = 37 * 18

37C2 = 666 ways

Hence the total number of ways that the 5 committees can be selected is expressed as 10C3 * 37C3 = 120 * 666

120 * 666 = 79,920 ways

answer: evaluate the objective function at each vertex to determine which - and -values, if any, maximize or minimize the function. example: find the minimum value and maximum value of the objective function f ( x , y ) = 4 x + 5 y , subject to the following constraints

use formula for the circle a=3.14

step-by-step explanation: