Combinations
Jane likes to wear 3 bracelets, each a different color. If she has 5 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 = 5 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:
| 5C3 = | 5! |
| 3!(5 - 3)! |
Calculate the numerator n!:
n! = 5!
5! = 5 x 4 x 3 x 2 x 1
5! = 120
Calculate the first denominator (n - r)!:
(n - r)! = (5 - 3)!
(5 - 3)! = 2!
2! = 2 x 1
2! = 2
Calculate the second denominator r!:
r! = 3!
3! = 3 x 2 x 1
3! = 6
Now calculate our combination value nCr for n = 5 and r = 3:
| 5C3 = | 120 |
| 2 x 6 |
| 5C3 = | 120 |
| 12 |
5C3 = 10
Therefore, there are 10 unique ways to choose 3 different color bracelets from 5 total bracelets
