Question

A box has 15 ping pong balls in it, numbered 1 through 15. How many distinct sets of four different numbers can be chosen from the box of ping pong balls if the order in which the numbers are chosen is not important?

Answer 1

Answer: 1365 combinations

Explanation:

What you're looking for if the order doesn't matter is the number of combinations.

You can use this formula for it:

`(n!)/(r!(n-r)!) \\`

Where n is the number of objects there are and r is the number of objects you take out.

We'll be using the variables:

n = 15

r = 4

Plug them in and solve:

`\\(15!)/(4!(15-4)!) \\\\= (15!)/(4!* 11!) \\\\= (1307674368000)/(24 * 39916800) \\\\= (1307674368000)/(958003200) \\= 1365`

Answer 2

1365

Explanation:

The number of ways in which you can choose r items as a set from n available items is given by n choose k which can be represented as
Here n = 15, r = 4

C(n, r)

C(n,r) = C(15,4)

The formula is

`C(n,r) = \binom{n}{r} = (n!)/(( r! (n - r)! ))`

where ! represents factorial of a number

n! = n x (n-1) x (n-2) x.......x 3 x 2 x 1

So

`C(15,4) = = (15!)/(( 4! (15 - 4)! ))\\\\ = (15!)/(4! * 11! )\\\\\\= 1365`