### Video Transcript

Let π 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 is the set containing the elements π and π, where π and π are elements of the
set π₯ and π is not equal to π. Determine the value of π of π,
where π of π is the number of elements in π.

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

But of course, choosing 10 and 11
would be the same as choosing 11 and then 10. And so weβre interested in
combinations. And π of π is seven choose
two. The formula of π choose π, the
number of ways of choosing π items, from a set of π is π factorial over π
factorial times π minus π 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 π of π, which was the
number of ways of choosing two unique elements from a set of seven where order
doesnβt matter, is 21. We can also apply this formula more
than once to help us answer counting problems.