Combinations
Jane likes to wear 3 bracelets, each a different color. If she has 8 bracelets each of a different color, how many combinations of 3 different-colored bracelets can she select to wear?Combination problems involve choosing r combinations from n items. In this case, n = 8 bracelets and r = 3 bracelets
The formula for a combination of choosing r unique ways from n possibilities is:
| nCr = | n! |
| r!(n - r)! |
where n is the number of items and r is the unique arrangements.
What is n!, r!, (n - r)!?
n! signifies a factorial. n!, for example is shown in our factorial lesson as being n! = n * (n - 1) * (n - 2) * .... * 2 * 1Plugging in our factorial numbers, we get:
| 8C3 = | 8! |
| 3!(8 - 3)! |
Calculate the numerator n!:
n! = 8!
8! = 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1
8! = 40,320
Calculate the first denominator (n - r)!:
(n - r)! = (8 - 3)!
(8 - 3)! = 5!
5! = 5 x 4 x 3 x 2 x 1
5! = 120
Calculate the second denominator r!:
r! = 3!
3! = 3 x 2 x 1
3! = 6
Now calculate our combination value nCr for n = 8 and r = 3:
| 8C3 = | 40,320 |
| 120 x 6 |
| 8C3 = | 40,320 |
| 720 |
8C3 = 56
Therefore, there are 56 unique ways to choose 3 different color bracelets from 8 total bracelets
