# Combinatorics intro problem 1 - solution

Finally, a post to end the tension about the unsolved problem! Have you been trembling with anticipation? If not, you still have a few seconds. [Insert obligatory trembling]

So if you remember, in our last post, we gave you this problem (and another one, but that's for a future post) to chew on:

You have three opaque boxes on a table. It is known that one of them contains two white balls, another has two black balls, and the last has one black and one white ball. The boxes have labels "2 white", "black and white", and "2 black", but you know that the muppet who put them on got them ALL wrong (so none of the boxes have the right label on them!). You are allowed to pick just one ball (without looking) from one of the boxes. Can you work out which boxes should have which labels?

Or maybe yours was a little better drawn

If you thought it through and drew a picture, you might have got something like this (on the right)

You can see the boxes have labels, but they're all wrong. But remember we can only see the labels!

So we have to make a choice - where should we draw the ball from? Does it matter? Here comes the first thing about Maths Olympiad (and all maths in general!)

There might be one "right answer", but there are always many ways to get there!

You might have done this with lots of cases - for example... "Case 1: if I pick the ball out of the box labelled '2 black' .... Case 2: if I pick the ball out of the box labelled '2 white' .... and so on"; you might have drawn out all the different possible ways the boxes could be labelled. If you get the right answer and you don't break maths while doing it, there's nothing wrong with it - it's a valid solution! You won't believe how much real maths is done this way.

But what makes maths cool is that there's often really nice, really clever ways to do things. I'll go over a nice solution from here on, so if you still want a solid shot at the problem,

Let's start by drawing a few situations. You might notice after a while that the box labelled 'black and white' must always contain either all black or all white balls. This is clear in hindsight because we know that all the boxes are wrongly labelled. But you'll be surprised - this is the key to the solution!

Maths tip number 1:

Try stuff out - you might notice interesting patterns that could help with a solution

This is just for example - if we drew a white ball, the box must really be 2W

So our strategy is to pick the ball out of the box labelled 'black and white'. If we get a black ball, we know that the box must contain two black balls, and if we get a white one, we know that the box must contain two white ones - because this particular box is guaranteed to never contain a mixture.

Why does that help us? Well... the ball we draw immediately betrays the true nature of that box.

Now, let's consider the two cases (I'll put true identities in bold from now on):

1. If we drew a black ball, the box labelled 'black and white' must really be '2 black'.
But then - what about the box labelled '2 white'? It can't be '2 white' because the labels always lie, and it can't be '2 black' because that one is already taken. That leaves us with 'black and white'.
Finally, the last box labelled '2 black' must then be '2 white' because that's the only box left. We're done! Just with pure logic!
It's the same with the second case...
2. If we drew a white ball, the box labelled 'black and white' must be '2 white'.
This forces the box labelled '2 black' to really be 'black and white' because it can't be '2 white' or '2 black', and this finally leaves the last box labelled '2 white' to really be '2 black'.

Hold on. We're done already? Yes! We've answered the question - we've provided instructions on how we would go about fixing the labels, after drawing that one ball. We've also covered all the possible cases (we could either draw a black or white ball - there's no random blue ball that's going to pop up), so we're done!

Maths tip number 2:

Don't be afraid to try case-by-case work. There might be nice ways around it, but cases are often a good way to finish a problem. Just a warning though: don't leave cases out!

For example, if you're splitting cases into x < 0 and x > 0, don't leave out the case when x = 0.

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Did you follow that? Was that too easy or hard? Please leave feedback, questions, corrections (or hate :'( ) in the comments below! We want to hear from you :)

BC

P.S. "combinatorics", often called "combi", is pronounced "com-bin-a-TOR-ricks", not "combine-a-torics"