Question Video: Evaluating Combinations by Using Properties of Combinations Mathematics

Write π‘›πΆπ‘Ÿ + 𝑛𝐢(π‘Ÿ + 1) in the form π‘ŽπΆπ‘.


Video Transcript

Write 𝑛 choose π‘Ÿ plus 𝑛 choose π‘Ÿ plus one in the form π‘Ž choose 𝑏.

Initially, we might look at these combinations and think we should expand them with the factorial formula to add them together. And while it’s possible to use this method, there’s a much more straightforward approach. Instead, we can use the recursive property for combinations. This tells us that 𝑛 minus one πΆπ‘Ÿ plus 𝑛 minus one πΆπ‘Ÿ minus one is equal to 𝑛 choose π‘Ÿ.

Now at this point, we’re looking at our combination and thinking, how does this help us since we don’t have any instances of 𝑛 minus one in our sum of combinations? But we know that we’re looking for something in the form π‘ŽπΆπ‘. So first, we convert our recursive property to include π‘Žβ€™s and 𝑏’s in place of the 𝑛’s and π‘Ÿβ€™s. And we have π‘Ž minus one choose 𝑏 plus π‘Ž minus one choose 𝑏 minus one is equal to π‘Ž choose 𝑏. We can then use this to try and work backwards.

Notice that in the recursive property, we’re adding two combinations with the same set size. That’s π‘Ž minus one. And in one combination, we’re choosing a certain value, that’s 𝑏, and in the other combination, we’re choosing one less than that, 𝑏 minus one. So now going back to our sum of combinations, we have π‘Ÿ plus one and π‘Ÿ, which is one less than π‘Ÿ plus one. And since π‘Ÿ is one less than π‘Ÿ plus one, we swap the terms in the recursive property around so that our π‘Ÿ corresponds to 𝑏 minus one. And our π‘Ÿ plus one corresponds to 𝑏.

Since we’re dealing with addition, it’s fine to change the order and the sum is still equal to π‘Ž choose 𝑏. Doing all this gives us 𝑛 equal to π‘Ž minus one. Now if 𝑛 equals π‘Ž minus one, adding one to both sides gives 𝑛 plus one equals π‘Ž. So with π‘Ž equal to 𝑛 plus one and 𝑏 equal to π‘Ÿ plus one, we have π‘Ž and 𝑏 in terms of 𝑛 and π‘Ÿ. And this means we’ve been able to write the two combinations in our sum as a combination in the form π‘Ž choose 𝑏. So we have 𝑛 choose π‘Ÿ plus 𝑛 choose π‘Ÿ plus one is equal to 𝑛 plus one choose π‘Ÿ plus one.

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