Problematic little chairs

Probability Level 3

There are $2016$ chairs in a row and two available colours, blue and red . On how many ways can we paint all chairs, so that number of blue chairs is even ?

(e. g. A case in which the first two chairs are blue and other are red, is different from the case in which the last two chairs are blue and other are red)

In solution section "inspiration" for the problem is mentioned.

2016!

2^{2016} - 2016

2^{2015}

2^{2016}

\frac{2016!}{1008!}

\left( \begin{matrix} 2016 \\ 1008 \end{matrix} \right)

1 solution

Milan Milanic
Jan 18, 2016

I didn't want to post this problem for a long time, because it is not mine. It is taken from math competition. The solution is so simple and logical, but at that time, I couldn't see it. Because of its "fancy" solution and a little chat with one brilliant member, I decided to post it anyway.

Solution:

The 1st chair can be painted either red or blue (2 choices). It is the same with the 2nd chair (2 choices), and the 3rd, the 4th, .... and so on. The 2015th can also be painted either red or blue (it doesn't matter, it is like the condition "even number of blue chairs" doesn't exist).

However, after choosing colours for the previous 2015 chairs, the 2016th chair's colour is determined (only one choice). Until now, 2015 chairs are already painted, and the number of blue chairs is either even or odd. If it is the first one, then 2016th chair will be red, on the other hand, if it is the latter one, the chair will be painted blue.

Therefore, solution would be $\boxed{2^{2015}}$ .

Problematic little chairs

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