### Video Transcript

If πP15 is equal to 23 times π
minus one P14, find π.

We have an equation with two
different permutations on either side. On the left, our set is size π,
and on the right, we have π minus one. On the left, weβre choosing 15, and
on the right, weβre choosing 14, which is 15 minus one. We actually have a property of
permutations that fits this pattern. It tells us for πPπ, itβs equal
to π times π minus one Pπ minus one. In our question, we have π, π
minus one, then π, π minus one. And this means the value of π will
be equal to the coefficient of this other permutation, in our case, 23. And therefore, we can say that π
equals 23. But you might be wondering, what if
you didnβt remember this property? Is there another way to solve?

If we know that we calculate πPπ
by taking π factorial over π minus π factorial, on the left we have π factorial
over π minus 15 factorial. And on the right, we have 23 times
π minus one factorial over π minus one minus 14 factorial, where π minus one is
in the π position and 14 is in the π position. We can do a bit of simplifying on
the right so that we have 23 times π minus one factorial over π minus 15
factorial. Since we have π minus 15 factorial
in the denominator on both sides, we can multiply both sides of the equation by π
minus 15 factorial, which will cancel out these terms. And then, we have π factorial is
equal to 23 times π minus one factorial.

But we also know the definition of
π factorial. And that means weβll substitute for
π factorial π times π minus one factorial. We now have an π minus one
factorial on both sides of our equation. So, we divide both sides of our
equation by π minus one factorial. And that term cancels out on both
sides, leaving us with π equals 23.