### Video Transcript

Determine the value of 66πΆ21 plus
the summation from π equals one to 15 of 81 minus ππΆ20.

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 15πΆ20. 81 minus 15 is 66, and then weβve
changed the upper bound of the summation to 14. And then, we see 66πΆ21 plus
66πΆ20, which should remind us of the recursive relationships in combinations that
tells us π minus one πΆπ plus π minus one πΆπ minus one equals ππΆπ. In this case, 66πΆ21 would be the
π minus one π and 66πΆ20 would be π minus one π minus one, which means when
these two values are combined, they equal 67πΆ21.

When we bring down the rest of the
summation, we consider what would happen if we plug in 14 to 81 minus π. 81 minus 14 is 67. This means we can pull out the term
67πΆ20 from our summation. And then again, this recursive
relationship would apply such that these two combinations can be simplified to
68πΆ21, 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 π and see that our final combination
for this recursion would be 80πΆ20.

Notice that this summation is from
π equals one to π 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 80πΆ21. We complete this relationship one
final time, 80πΆ21 plus 80πΆ20, which simplifies to 81πΆ21.