It occurred to me today that I could ‘meaningfully’ introduce Pascal’s triangle before students meet the binomial theorem.

When teaching basic set theory, the most interesting part for me is to ask students to investigate the number of subsets of a set with n elements. For example, the set {a, b} has four distinct subsets – {}, {a}, {b}, {a, b}. See the table below for more data:

If you haven’t investigated this before, you’ll notice that a set with n elements has 2^n subsets (always a lovely link to exponential expressions/sequences/functions within the syllabus).

It’s common knowledge for those who love Pascal’s triangle that row n of Pascal’s triangle sums to 2^n.

Since position r, row n, of Pascal’s triangle represents n objects choose r of them, it is clear that this is equivalent to a set with n elements. E.g. 4 elements, choose 2 of them and there’d be 6 combinations. Indeed, this helps us show why the number of subsets of an n element set is 2^n by expanding (1 + 1)^n.

The illuminating part of this is that, if a student wants to know how many subsets with two elements exist from a set with 4 elements {a, b, c, d} , they simply need to generate the 4th row of Pascal’s triangle and take a look at the 2nd element (6 – {a, b}, {a, c}, {a, d}, {b, c}, {b, d}, {c, d}). Without even listing the number of subsets which have 4 elements in from a set with 7 elements, you can just generate the seventh row of Pascal’s triangle and see that there’d be 35 distinct subsets.

So not only is there a connection between set theory and exponential functions, there’s also a look into Pascal’s triangle without even needing to mention combinations if you didn’t think it appropriate. They just generate the triangle, observe that each row sums to 2^n just like the number of subsets of a set, and then convince themselves that each element in row n, position r, corresponds to the number of subsets of an n element set with r elements.

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