Video: Using the Recursive Rule of Combinations to Simplify a Sum of Combinations

By applying the relation 𝑛Cπ‘Ÿ + 𝑛Cπ‘Ÿ βˆ’ 1 = 𝑛 + 1Cπ‘Ÿ, deduce the value of 59C2 + 59C3.


Video Transcript

By applying the relation 𝑛 choose π‘Ÿ plus 𝑛 choose π‘Ÿ minus one equals 𝑛 plus one choose π‘Ÿ, deduce the value of 59 choose two plus 59 choose three.

This is called the recursive relationship. And it can help us to simplify expressions. In this case, we’re looking to deduce the value of 59 choose two plus 59 choose three. So let’s compare this expression with the recursive relationship. In this relation, we see that on the left-hand side, we have two terms with the same value of 𝑛. We have 59 choose two and 59 choose three. So we’re going to let 𝑛 be equal to 59. Then we have π‘Ÿ and π‘Ÿ minus one. The first term has a larger value of π‘Ÿ. And the second term has a value of π‘Ÿ that’s one less.

So let’s imagine we’re going to swap 59 choose three and 59 choose two around so that it matches this criteria. Then we see π‘Ÿ must be equal to three. π‘Ÿ minus one is three minus one, which is two, as we require. According to the relation given in the question then, 59 choose three plus 59 choose two must be equal to 𝑛 plus one, so 59 plus one choose three. That’s 60 choose three. So to deduce the value of the expression in our question, we need to work out what 60 choose three is.

And so we recall that 𝑛 choose π‘Ÿ is 𝑛 factorial over π‘Ÿ factorial times 𝑛 minus π‘Ÿ factorial. Now, we mustn’t confuse 𝑛 and π‘Ÿ with the values we defined earlier. This time, 𝑛 is going to be equal to 60, and π‘Ÿ is still equal to three. This gives us that 60 choose three is 60 factorial over three factorial times 60 minus three factorial or 60 factorial over three factorial times 57 factorial. Now, we could use our calculator to evaluate this, but let’s look at how the definition of the factorial can help us simplify a little.

𝑛 factorial is 𝑛 times 𝑛 minus one times 𝑛 minus two all the way down to one. This means 60 factorial is 60 times 59 times 58 and so on. But of course, we see that we can actually rewrite that further as 60 times 59 times 58 times 57 factorial. This means our expression for 60 choose three can be rewritten as shown. And this is great because we can now divide through by a constant factor of 57 factorial. In fact, three factorial is equal to six, so we can also divide our numerator and denominator by six. And we see that 60 choose three is 10 times 59 times 58 over one. 59 multiplied by 58 is 3,422. So 10 times 59 times 58 is 34,220. 59 choose two plus 59 choose three is, therefore, 34,220.

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