# Thread: 3 eggs 1 Box problem

1. ## 3 eggs 1 Box problem

I heard this problem explained on the radio today and was wondering if anyone could work out the answer.

You have an 6 slot egg box with 3 eggs in it, what are the number of possible
combinations you can put the eggs inside, without removing any eggs from the box ( and all the eggs being identical, so basically each slot can either have an egg in it or not).

I'm not very good at maths but my first thought was to try doing 2 to the power of 6, then I realised there was only 3 eggs so not every slot can be filled with eggs so that wouldn't work. I then started thinking of it kind of like a 6 digit binary number, but with the limit that it must always have 3 zeroes in it ( because 3 slots will always be empty).
In the end i kind of thought that the number was obviously below 64 but definitely above 27, and thats about as far as i could get with it.
An answer would be greatly appreciated!

2. Originally Posted by armlesscorps
You have an 6 slot egg box with 3 eggs in it, what are the number of possible combinations you can put the eggs inside, without removing any eggs from the box ( and all the eggs being identical, so basically each slot can either have an egg in it or not).
Basically we assume that the 'slots' are distinct and the eggs are identical.
So any arrangement of the string $111000$ will model the eggs inside.
That answer is $\frac{6!}{(3!)^2}$.

3. ah awesome thanks a lot. so if I wanted to work out what would happen in an eggbox with 16 slots and 8 eggs using the same rules, I would divide the 16 factorial by the 8 factorial squared?

Also, if there were 6 slots and I could remove eggs from the box this time ( so basically a 6 digit binary). The answer would be the possible arrangements of the string 111111? How would that work out mathematically and could it be worked out using factorials again?

by the way when I said I'm not good at maths, I had to google to find out what ! meant, just to let you know