### Video Transcript

If the ratio of π choose nine to π choose seven is seven to nine, find π.

This question involves the ratio of binomial coefficients π choose nine and π choose seven. And if weβd like to, we could write this equation of ratios as an equation of quotients. π choose nine divided by π choose seven is seven-ninths. And as before, we wanted to use this to find the value of π. This quotient of binomial coefficients on the left-hand side might remind us of a rule that we know. π choose π divided by π choose π minus one equals π minus π plus one all over π. But if we look closer, itβs not a perfect match.

To get the numerator to be π choose π, we have to substitute π equals nine. We can make this substitution and simplify to find the quotient of π choose nine and π choose eight. But we wanted π choose seven in the denominator not π choose eight. If we instead substitute π equals eight, then the denominator is π choose seven as we want. But the numerator is now π choose eight which we donβt want. One of the quotients has the correct numerator. And the other has the correct denominator. The trick is to see that if we multiply these two quotients together, then the π choose eight cancel. And weβre left with π choose nine over π choose seven which is what we have here.

We can substitute π minus eight over nine for π choose nine over π choose eight and π minus seven over eight for π choose eight over π choose seven to get π minus eight over nine times π minus seven over eight equals seven-ninths. Now, if you donβt know this fact or if you donβt think that you would have been able to see how to apply it to this question by yourself, then donβt worry weβll see another method later on. But for the moment, letβs focus on solving this equation. We can multiply both sides by nine, cancelling the nines in the denominators and also by eight getting rid of the denominator eight on the left-hand side and giving us seven times eight which is 56 on the right-hand side.

Now, we have something recognisable as a quadratic equation. We expand the left-hand side to get π squared minus 15π plus 56. Subtract 56 from both sides getting π squared minus 15π equals zero. We take the common factor of π out of the left-hand side. And we see that this quadratic gives us two solutions π equals zero and π equals 15. But we canβt yet say that π equals zero and π equals 15 are both answers to our original question. We have to make sure that the values make sense in the context of the original question which asks us about the ratio of π choose nine to π choose seven.

Does it make sense for π to be zero here, could the ratio of zero choose nine to zero choose seven really be seven to nine? Well, no. Zero choose nine and zero choose seven which are, respectively, the number of ways of choosing nine objects from a set of zero objects and the number of ways of choosing seven objects from a set of zero objects are either both undefined or both zero. Either way their ratio is certainly not seven to nine. On the other hand, it does make sense to talk about 15 choose nine and 15 choose seven and their ratio. And so π equals 15 does make sense in the context of the original question and is, therefore, our answer.

You can check this if you have a calculator with a choose button. When I type 15 choose nine over 15 choose seven into my calculator and press enter, it does indeed give me seven-ninths. Alternatively I can verify this result by hand using the formula for π choose π. π choose π is π factorial over π factorial π minus π factorial. And so 15 choose nine over 15 choose seven is 15 factorial over nine factorial six factorial all over 15 factorial over seven factorial eight factorial. If you multiply numerator and the denominator by nine factorial six factorial seven factorial eight factorial we get 15 factorial seven factorial eight factorial in the numerator and 15 factorial nine factorial six factorial in the denominator.

The 15 factorials cancel. And we can write seven factorial as seven times six times five times four times three times two times one which is seven times six factorial. And similarly in the denominator, we can write nine factorial as nine times eight factorial. And we see that all the factorials cancel leaving seven over nine as required. This method of verification gives us another solution method which doesnβt involve remembering the expression for π choose π over π choose π minus one.

Letβs clear some room and see this. We start with the same quotient equation using our formula for π choose π to write π choose nine and π choose seven in terms of factorials. And as with the verification, we can simplify the left-hand side by multiplying by lots of factorials. Of course, any other methods to divide fractions including multiplying by the reciprocal of the fraction in the denominator would also give the same fraction on the left-hand side. We can see immediately that the π factorials cancel and we get more cancellation if we write nine factorial as nine times eight times seven factorial. And similarly, but slightly more tricky, is the fact that π minus seven factorial is π minus seven times π minus eight times π minus nine factorial.

With these substitutions, on the left-hand side we get seven factorial times π minus seven times π minus eight times π minus nine factorial over nine times eight times seven factorial times π minus nine factorial. And the seven factorials in the numerator and denominator cancel as do the π minus nine factorials. Tidying up, then we get the equation π minus seven times π minus eight over nine times eight equals seven-ninths which is basically the same as the equation we got before.

Certainly, if we multiply by nine times eight on both sides, we get π minus seven times π minus eight equals 56 which differs from this equation only in the order of the factors on the left-hand side. From here on then, the solution is the same as before. We solve the quadratic, eliminate these solution π equals zero, and conclude that our answer is that π is 15.

This method didnβt require us to use the formula for π choose π over π choose π minus one nor did we need the trick of writing π choose nine over π choose seven as π choose nine over π choose eight times π choose eight over π choose seven. In general, using the formula for π choose π more reliably gives a straightforward solution. And if for any reason you need the quotient of π choose π over π choose π minus one, you can drive it using this formula for π choose π.