Musings

Peas, Carrots, Beans

My daughter brought home the following problem from her 2^nd grade math class:

You have 10 pots in which to plant vegetables. You can plant either peas, carrots or beans in each pot. You could plant 7 peas, 2 carrots and 1 bean plant, or 3 peas, 5 carrots and 2 beans, or … as long as you have at least 1 plant of each type and a total of 10 plants. How many different combinations are there?

This is a wonderful problem … for a slightly older child. My daughter’s answer, “more than 10,” was arrived at by a really noble attempt at brute-force enumeration. For 2^nd graders, who have not yet mastered division, the actual solution is tantalizingly beyond their grasp.

Let’s solve a simpler problem first. Say we just want to plant two vegetables: peas and carrots. Line the 10 pots up in a row. We’ll plant all the peas on the left and the carrots on the right, and we’ll put a cardboard divider between them. There are 9 places to put the divider, so there are 9 different combinations of 2 vegetables that can be planted in the 10 pots. For 3 vegetables, we need to put in 2 dividers (peas on the left, carrots in the middle and beans on the right). There are 9 places to put the first divider and 8 places to put the second divider (since we’ve already taken one of the slots it could go in). That sounds like $9\times 8= 72$ possibilities. But we’ve overcounted, since swapping the locations of the two dividers gives us the same configuration of plants. So the correct answer is $9\times 8/2=36$ combinations. [More generally, with $N$ pots and $k$ types of vegetables, there are $\scriptsize{\left(\array{N-1\\ k-1}\right)}$ combinations.]

Maybe next year, I could actually explain this solution to her. Maybe, when she’s older still, she could come up with this solution on her own. But not this year …

So what does the teacher have in mind with this exercise?