EdExcel GCSE question about Hannah and the sweets: show that $n^2-n=90$ This is my reconstruction of the EdExcel GCSE question that has caused such a Twitter storm in the UK in the last 24 hours, along with its solution.

Hannah has a bag containing $n$ sweets, 6 of which are orange. She eats two sweets at random from the bag.
The probability that the two sweets Hannah eats are both orange is $\frac{1}{3}$. Show that $n^2 - n = 90$.

Possible follow-up (I don't know if this was part of the original): how many sweets were there in total in the bag?
 A: 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 $$6 \times 5 \times 3 = n(n-1)$$ or $$90 = n^2 - n$$ 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 $1$ 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 $10$ sweets or $-9$ sweets in the bag. Obviously there can't have been $-9$ sweets; so we conclude that there were $10$ sweets in the bag.
A: Here is an alternative solution:
The probability that the first two sweets selected from a bag are orange if six of the $n$ sweets in the bag are orange is 
\begin{align*}
\frac{\binom{6}{2}}{\binom{n}{2}} & = \frac{1}{3}\\
\frac{\frac{6 \cdot 5}{2 \cdot 1}}{\frac{n(n - 1)}{2 \cdot 1}} & = \frac{1}{3}\\
\frac{6 \cdot 5}{n(n - 1)} & = \frac{1}{3}\\
\frac{30}{n(n - 1)} & = \frac{1}{3}\\
90 & = n(n - 1)\\
90 & = n^2 - n\\
\end{align*}
To determine the number of sweets in the bag, we factor the equation $n^2 - n - 90 = 0$.
\begin{align*}
n^2 - n - 90 & = 0\\
(n - 10)(n + 9) & = 0
\end{align*}
Hence, $n = 10$ or $n = -9$.  Since the number of sweets in the bag cannot be negative, the number of sweets in the bag is $10$.
A: The chance of an orange sweet being chosen the first time is 6/n. Everyone got that right.
The chance of an orange sweet being chosen the second time is (6-6/n)/(n-1). Most people seem to have got that wrong.
If the chance of both sweets being orange is 1/3, then n X n=108, but that's impossible because n must be an integer. In fact the chance of both sweets being orange can never be 1/3. The question is in error either deliberately or by mistake.
