There are 20 students in a fraternity. In how many ways can a committee of 9 students be selected?

Question

There are 20 students in a fraternity. In how many ways can a committee of 9 students be selected?

 

The number of ways to form a 9-student committee is?

Answer 1

This is a combination of 20 things taken 9 at a time. In a combination the order does not make a difference.

 

Example: In a combination 1,2,3 and 2,3,1 and 3,1,2 and 3,2 1, and 1,3,2, and 2,1,3 are the same because they contain the same number. So in selecting the committee, the order in which the members does not matter. 

 

The formula for a combination

 

n!/[r!(n - r)!]

 

Now that looks more complicated than it is.

 

n! = n(n - 1)(n - 2)(n - 3) .......  So 4! = 4 · 3 · 2 · 1 

 

In the formula n = the number from which selections can be made. Here n = 20.

r = the number of selections being made. Here r = 9.

 

Putting the numbers in the formula:

 

20!/[9!(11!)]

 

This reduces to:

 

20 · 19 · 18 · 17 · 16 · 15 · 14 · 13 · 12 / 9 · 8 · 7 · 6 · 5 · 4 · 3 · 2 · 1   (If I could write a "real" fraction here I could make it look simpler.)

 

Now this calculates to:

 

167,960

 

With twenty students 167,960 committees of nine members can be made.

Answer 2

Hi Violet,

 

This is a combination problem

 

It is a combination of 20 things taken 9 at a time

 

our numerator would be 20*19*18*17*16*15*14*13*12

 

our denominator would be 9*8*7*6*5*4*3*2*1

 

simplify and then multiply

 

the final answer should be 167,960

 

Hope this helps

Answer 3

The number of ways to choose 9 students from 20 is given by 20 choose 9 = 20!/(9!*(20-9)!) = 20!/(9!*11!) = 167960