Question Video: Combination without Replacement Mathematics

Let 𝑋 = {π‘₯ : π‘₯ ∈ β„€, 10 ≀ π‘₯ ≀ 16} and π‘Œ = {{π‘Ž, 𝑏} : π‘Ž, 𝑏 ∈ 𝑋, π‘Ž β‰  𝑏}. Determine the value of 𝑛(π‘Œ), where 𝑛(π‘Œ) is the number of elements in π‘Œ.

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Video Transcript

Let 𝑋 be the set containing π‘₯, where π‘₯ is an integer greater than or equal to 10 and less than or equal to 16, and π‘Œ is the set containing the elements π‘Ž and 𝑏, where π‘Ž and 𝑏 are elements of the set π‘₯ and π‘Ž is not equal to 𝑏. Determine the value of 𝑛 of π‘Œ, where 𝑛 of π‘Œ is the number of elements in π‘Œ.

Let’s just begin by looking at what this set notation is actually telling us. Set 𝑋 contains the number π‘₯, which is an integer greater than or equal to 10 and less than or equal to 16. And so 𝑋 could be rewritten as the set containing the elements 10, 11, 12, 13, 14, 15, and 16. Then, we have set π‘Œ, and set π‘Œ contains all pairs of integer combinations π‘Ž and 𝑏 such that π‘Ž is not equal to 𝑏. But both π‘Ž and 𝑏 are elements of set 𝑋. So it’s like a list of pairs of numbers, and we need to work out how many pairs are in that list. And so 𝑛 of π‘Œ is the number of ways of choosing two elements from a total of seven. And that’s because there are seven elements in set 𝑋. Since π‘Ž cannot be equal to 𝑏, there’s no repetition.

But of course, choosing 10 and 11 would be the same as choosing 11 and then 10. And so we’re interested in combinations. And 𝑛 of π‘Œ is seven choose two. The formula of 𝑛 choose π‘Ÿ, the number of ways of choosing π‘Ÿ items, from a set of 𝑛 is 𝑛 factorial over π‘Ÿ factorial times 𝑛 minus π‘Ÿ factorial. And so seven choose two is seven factorial divided by two factorial times seven minus two factorial. Now, we can write seven minus two factorial as five factorial. And then we can also write the numerator seven factorial as seven times six times five factorial. Then we divide through by five factorial.

And since two factorial is two, we can also divide through by two. And seven choose two simplifies to seven times three over one, which is simply 21. And so 𝑛 of π‘Œ, which was the number of ways of choosing two unique elements from a set of seven where order doesn’t matter, is 21. We can also apply this formula more than once to help us answer counting problems.

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