Question Video: Using the Addition Rule of Combinations to Simplify a Sum of Combinations | Nagwa Question Video: Using the Addition Rule of Combinations to Simplify a Sum of Combinations | Nagwa

Question Video: Using the Addition Rule of Combinations to Simplify a Sum of Combinations Mathematics • Third Year of Secondary School

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Determine the value of 66𝐢₂₁ + (βˆ‘_(π‘Ÿ = 1)^(15) (81 βˆ’ π‘Ÿ)𝐢₂₀.

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Video Transcript

Determine the value of 66𝐢21 plus the summation from π‘Ÿ equals one to 15 of 81 minus π‘ŸπΆ20.

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 15𝐢20. 81 minus 15 is 66, and then we’ve changed the upper bound of the summation to 14. And then, we see 66𝐢21 plus 66𝐢20, which should remind us of the recursive relationships in combinations that tells us 𝑛 minus one πΆπ‘Ÿ plus 𝑛 minus one πΆπ‘Ÿ minus one equals π‘›πΆπ‘Ÿ. In this case, 66𝐢21 would be the 𝑛 minus one π‘Ÿ and 66𝐢20 would be 𝑛 minus one π‘Ÿ minus one, which means when these two values are combined, they equal 67𝐢21.

When we bring down the rest of the summation, we consider what would happen if we plug in 14 to 81 minus π‘Ÿ. 81 minus 14 is 67. This means we can pull out the term 67𝐢20 from our summation. And then again, this recursive relationship would apply such that these two combinations can be simplified to 68𝐢21, 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‘Ÿ and see that our final combination for this recursion would be 80𝐢20.

Notice that this summation is from π‘Ÿ equals one to π‘Ÿ 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 80𝐢21. We complete this relationship one final time, 80𝐢21 plus 80𝐢20, which simplifies to 81𝐢21.

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