# Permutations, Combinations, and Why?

We’ve spent a couple of days in precal talking about permutations and combinations.

The first day, we did a fairly straight-forward permutations/combinations lesson. How many ways can you arrange 5 cars in a driveway? How many distinguishable arrangements of the letters in the your full name are there? (Most people have at least one repeated letter.) How many 3-letter “words” can you form if you never repeat a letter? If you only use consonants? If you only use vowels? If I have ten books and want to take 3 on a trip, how many different sets of 3 do I have to choose from?

It was a fairly standard lesson. We’d arrange 2 cars, then 3, then they’d arrange 4 cars, then we’d find the pattern and make predictions. What was slightly different (and what I want to do more of) is this: when we found the pattern, I’d make them figure out why. If they couldn’t, I’d help, but they had to try on their own first.

Why is the number of ways you can arrange 4 cars equal to 4*(the number of ways you can arrange 3 cars)? Why would we divide by the factorial of the number of times a letter is repeated in your name? Repeat with each different topic. When we hit permutations and combinations, I introduced the vocabulary, as well as the notation ${n \choose k}$ and the phrase “binomial coefficient.”

Today, we did Pascal’s Triangle. We started by expanding binomials: $(x+y)^0, (x+y)^1, (x+y)^2, (x+y)^3,$ … . I wrote the coefficients in a triangle, they saw the pattern, and we expanded more binomials. Then I had them compute binomial coefficients, and we arranged them in a triangle. They caught on pretty quickly to the fact that the binomial coefficients we’d computed on Wednesday were, in fact, the coefficients of expanded binomials.

Why would it be true that the coefficient of $x^3y$ in the expansion of $(x+y)^4$ would be the same as the number of ways to choose 3 items out of 4?