Arby's Algebra

Arby's has recently had a commercial announcing their current deal, where you can choose 4 items from a list of 8 choices. They mention that this gives you 330 different options. Counting up this total is a nice example of combinatorics, so let's see how to figure out the total value.

Before getting into the math, it's helpful to assign a number to each of the eight menu choices. Rather than talking about an order as sandwich, fries, drink, apple turnover, it's simpler to call this 1,2,3,4. In the items below, when something like item 1 is mentioned, that just means I'd rather type "1" than "roast beef sandwich".

There are some approaches that are reasonable, but that don't give the right answer. First, you might consider that you have 8 choices for each of your 4 items, and simply count up 8*8*8*8, which gives 4096. This gives the total number of four-digit numbers where all the digits are from 1 to 8. The problem is that this double-counts some orders: the orders 1,1,2,3, 1,2,3,1, 3,2,1,1 are all counted as separate orders, when really they are the same.

Another approach is the 8 choose 4 method. The logic of this is that we have 8 items and we will choose 4 of them. There are 8 choices for the first item; once that's selected, there are 7 remaining choices for the next item, 6 for the third item, and 5 for the last. The total is 8*7*6*5 =1680. There are two problems with this. The first is that we're again double-counting some orders; this method will include both 1,2,3,4 and 4,3,2,1 as separate orders. We could solve that by dividing out these repetitions, but the second problem is worse: we're missing some orders. This method doesn't include repeating one item more than once. An order like 1,1,2,3 or 4,4,4,4 (for someone who loves apple turnovers) is never counted.

The correct way to look at this is as a "balls in buckets" problem (technically, combinations with repetition). The 8 menu items are the buckets, and we have 4 balls (choices in our order) to place in these buckets. If we place 2 balls in bucket 1, a ball in bucket 2, and a ball in bucket 3, that's an order of 1,1,2,3. All balls in bucket 4 is an order of 4,4,4,4. Now, here's the trick:

We can picture this as 7 dividers separating our 4 items into the 8 buckets. Visually, this means we translate our orders this way:

Order 1,2,3,4: o/o/o/o////
Order 4,4,4,4: ///oooo////
Order 2,4,5,8: /o//o/o///o

With the orders written this way, using the o and / characters, what we want to know is: how many different ways can we arrange 7 /s and 4 os? For this, we want an 11 choose 4 calculation. We have 11 total characters, and we need to choose 4 of them to be o's.

To start this calculation, we use 11*10*9*8, just like in the 8 choose 4 example above. As mentioned there, this has the problem of double-counting; it counts placing os in spots 1,2,3,4 as well as placing them in spots 4,3,2,1. This means that to get our final answer, we need to divide out these double-counts.

Luckily, that's pretty straightforward. The number of double-counts is just the number of ways to rearrange four numbers, and by the same logic we know this is 4*3*2*1. The final calculation is:

11*10*9*8 / 4*3*2*1
= 7920 / 24
= 330

Wikipedia link: Combinations