Question Video: Evaluating Combinations to Find an Unknown Then Evaluating This Value in a Combination | Nagwa Question Video: Evaluating Combinations to Find an Unknown Then Evaluating This Value in a Combination | Nagwa

# Question Video: Evaluating Combinations to Find an Unknown Then Evaluating This Value in a Combination Mathematics • Third Year of Secondary School

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Given that πCβ = 2π, evaluate πCβ.

03:25

### Video Transcript

Given that π choose three is equal to two π, evaluate π choose three.

Letβs begin by recalling what we actually mean by this notation. π choose π is the number of ways of choosing π items from a total collection of π items when order doesnβt matter. Itβs a combination. π choose π or πCπ is evaluated by dividing π factorial by π factorial times π minus π factorial. This means we can form an expression for π choose three. By letting π be equal to three, we get π factorial over three factorial times π minus three factorial.

Weβre told this is equal to two π. So letβs form an equation in π. We get π factorial over three factorial times π minus three factorial. But, of course, by the definition of the factorial, we know π factorial is π times π minus one times π minus two, and so on. This means we can write it as π times π minus one times π minus two times π minus three factorial.

And so, if we rewrite our numerator in this form, we see that we can divide through by a constant factor of π minus three factorial such that π times π minus one times π minus two over three factorial is equal to two π.

Letβs solve this equation for π by first multiplying through by three factorial. Now, three factorial is three times two times one. So itβs actually six. And we get π times π minus one times π minus two equals 12π. Next, weβre going to distribute our parentheses. We get π times π squared minus three π plus two equals 12π. Then we distribute further to get π cubed minus three π squared plus two π equals 12π.

Now, we have a cubic. So weβre going to subtract 12π from both sides to get this equal to zero. And we get π cubed minus three π squared minus 10π equals zero. Which we can solve for π by factoring first by removing the π and then factoring the quadratic π squared minus three π minus 10 to get π plus two times π minus five.

The roots of this equation are the values of π such that π is equal to zero, π plus two is equal to zero, or π minus five is equal to zero. Thatβs π equals zero, π equals negative two, and π equals five. Now, in fact, π must be a positive integer. So we disregard two of the solutions, and we get π is equal to five.

And weβre now able to evaluate π choose three. In this case, itβs actually five choose three. Going back to our definition for π choose π, weβre going to let π be equal to five and π be equal to three. That gives us five factorial over three factorial times five minus three factorial, which is five factorial over three factorial times two factorial.

This time, weβre going to write our numerator as five times four times three factorial. Allowing us to divide through by a constant factor of three factorial. But also two factorial is just two. So weβre going to divide through by two. And we see that five choose three is five times two, which is simply 10.

And so given that π choose three is equal to two π, we find π choose three is equal to 10.

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