Question

In how many ways can a committee of 4 be chosen from a group of 9 people?

Answer 1

Since this is a combination not a permutation problem, (order does not matter) you should use the "n choose k" formula.

C=n!/(k!(n-k)!)  where C is the number of unique combinations, n equals the total number of possible choices and k equals the specific number of choices.  In this case:

C=9!/(4!(9-4)!)

C=9!/(4!5!)

C=362880/(24*120)

C=362880/2880

C=126

So there are 126 unique ways to pick 4 people from a group of 9 people.

Answer 2

Answer:

126 ways.

Step-by-step explanation:

Since the order does not matter, the solution is obtained through a combination where we choose "r of n", where r is the amount of things we choose and n the total number of things that can be chosen.

In the given case,

r = 4

n = 9

The combinations form uses factorial numbers. This is the formula:

$nCr=\frac{n!}{(n-r)! r!}$

The factorial function (symbol:!) means that descending numbers are multiplied to 1.

We substitute the values in the equation and get

$9C4=\frac{9!}{(9-4)! 4!}$

$9C4=\frac{9.8.7.6.5!}{5! 4!}$

$9C4=\frac{9.8.7.6}{4!}$

$9C4=\frac{9.8.7.6}{4.3.2}$

$9C4=\frac{3024}{24}$

9C4 = 126

Therefore, there are 126 ways to choose a committee of 4 from a group of 9 people.

Hope this helps!