Video Transcript
Determine the value of 66𝐶21 plus
the summation from 𝑟 equals one to 15 of 81 minus 𝑟𝐶20.
At first, this expression looks
like it would be incredibly difficult to evaluate or to simplify. However, by looking closely at the
summation, where the upper bound is 15, we can remove one of the combinations 81
minus 15𝐶20. 81 minus 15 is 66, and then we’ve
changed the upper bound of the summation to 14. And then, we see 66𝐶21 plus
66𝐶20, which should remind us of the recursive relationships in combinations that
tells us 𝑛 minus one 𝐶𝑟 plus 𝑛 minus one 𝐶𝑟 minus one equals 𝑛𝐶𝑟. In this case, 66𝐶21 would be the
𝑛 minus one 𝑟 and 66𝐶20 would be 𝑛 minus one 𝑟 minus one, which means when
these two values are combined, they equal 67𝐶21.
When we bring down the rest of the
summation, we consider what would happen if we plug in 14 to 81 minus 𝑟. 81 minus 14 is 67. This means we can pull out the term
67𝐶20 from our summation. And then again, this recursive
relationship would apply such that these two combinations can be simplified to
68𝐶21, and then we bring down the rest of our summation. This means we can take this
recursion and do this operation again and again using the upper bound of the
summation until we got to one. Without writing out all of these
steps, we can plug in one into our 81 minus 𝑟 and see that our final combination
for this recursion would be 80𝐶20.
Notice that this summation is from
𝑟 equals one to 𝑟 equals 13. We need to use this recursive
relationship property 12 more times, so we get 68 plus 12 is 80. And we can then say that following
this recursive relationship, we would have 80𝐶21. We complete this relationship one
final time, 80𝐶21 plus 80𝐶20, which simplifies to 81𝐶21.
