Finding combinations

Probability is not the easiest subject in math…but it is incredibly useful so it can definitely be worth the time to understand it.

Here’s a problem you may find in precalculus:

You’ve got 6 baseball bats of varying lengths and weights but only room for 3 of them on your road trip. How many different combinations of bats can you take with you?

For the first choice you have 6 options, the second choice you have 5 options and the third choice you have 4 options. That’s 6x5x4 combinations or 120 combinations.

However, it doesn’t matter what order you put the bats in your bag. As it stands, you are overcounting the number of combinations. For example, you are counting Bat A, Bat B, and Bat C as one combination and Bat B, Bat C, Bat A as another combination. To avoid this, make sure to divide by the different orders you can have. For the first slot, you can choose 3 different bats, the second slot has the option of 2 different bats and the third slot has just 1 final choice. Thus, you must take 120 and divide it by 3x2x1 (or 6) which is 20.

Using the formal formula “n choose k” (in this case, n=6 and k=3) the problem will use factorials and look something like this:

6!/3! x 1/3!= 20.

On your TI-83 plus or TI-84 plus, it will look like:

You can find “nCr” under the MATH button if you scroll to the right and hit PRB, then 3.