### Video Transcript

If π choose four is equal to 35, find π choose π minus two.

Remember, π choose π, sometimes pronounced πCπ and written as shown, is π factorial over π factorial times π minus π factorial. By replacing π with four, we find π choose four to be equal to π factorial over four factorial times π minus four factorial. But of course, the question tells us this is equal to 35. So letβs create an equation. We get π factorial over four factorial times π minus four factorial equals 35. Letβs multiply both sides of this equation by four factorial, where four factorial is four times three times two times one or 24.

When we do, we get π factorial over π minus four factorial equals 840. But by the definition of the factorial, π factorial is π times π minus one times π minus two and so on. We can write this as π times π minus one times π minus two times π minus three times π minus four factorial. And this means we can divide through by a factor of π minus four factorial. And weβre left with an equation. Itβs π times π minus one times π minus two times π minus three equals 840. To solve this for π, weβre going to need to distribute our parentheses and set it equal to zero.

Weβll begin by multiplying π minus two by π minus three to give us π squared minus five π plus six. Then we multiply that quadratic expression by π minus one to give us π cubed minus six π squared plus 11π minus six. And then we multiply that cubic by π. And our equation is now π to the fourth power minus six π cubed plus 11π squared minus six π equals 840.

Letβs set this equal to zero by subtracting 840 from both sides. And so our equation is π to the fourth power minus six π cubed plus 11π squared minus six π minus 840 equals zero. So how do we solve for π? Well, we could use the factor theorem and fully factor this quartic expression. Alternatively, we can use a polynomial solver on our calculator. Either way, we get two real solutions and two complex solutions. Now, in the definition for π choose π, π must be a real integer. So weβre interested in these two solutions π is equal to seven or π is equal to negative four.

In fact, π must be positive. And so weβre going to choose this solution and disregard any further ones. And now that we know the value of π, we need to work out π choose π minus two. Thatβs seven choose seven minus two, which is seven choose five. Going back to our definition of the factorial, we see this is equal to seven factorial over five factorial times seven minus five factorial.

Letβs write seven factorial as seven times six times five factorial. We can also write seven minus five as two. And now, we see we can divide through by a constant factor of five factorial. But two factorial is also two. So weβre going to divide the numerator and denominator by two. And we see weβre left with seven times three, which is equal to 21. And so if π choose four equals 35, π choose π minus two must be equal to 21.