If both sweets that Hannah eats are orange, then of course the first one must be orange, and so must the second one. We can calculate the probability of that happening.

For the first sweet, there are $n$ in total, and 6 are orange, so the probability that the one she chooses is orange is $\frac{6}{n}$.

Once she's done this, there are now $n-1$ sweets left, 5 of which are orange. So the probability that the second one is also orange is $\frac{5}{n-1}$.

The overall probability (that both of these things happen) is then the product of these: $\frac{6}{n} \times \frac{5}{n-1}$.

Now, we are told in the question that the probability that they are both orange is $\frac{1}{3}$. So we have $$\frac{6}{n} \times \frac{5}{n-1} = \frac{1}{3}$$

If we multiply both sides by $n$, then by $n-1$, then by 3, we get $ \times 5 \times 3 = n(n-1)$$ or $ = n^2 - n$$ which is what we were asked to show.

As a follow-up, we can find the value of $n$, i.e., the total number of sweets in the bag. We need to solve $$n^2 - n - 90 = 0$$ which we can do by factorising: $$(n-a)(n-b) = n^2 - n - 90$$

We need two values whose sum is $ and whose product is $-90$. It is not hard to see that the values $a=10$ and $b=-9$ will do.

This means that there were either $ sweets or $-9$ sweets in the bag. Obviously there can't have been $-9$ sweets; so we conclude that there were $ sweets in the bag.

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