Monday, 30 May 2016

How many photographs (permutations including zero)

My wife and I have recently had our third child - hence I've not been blogging much lately.  However, I've been thinking about blog ideas, or more specifically, I've been thinking of mathematical investigations that I could explore and then, if they were interesting, share on my blog.

Our house is full of family photographs - almost every room has family photographs in it - and in particular on one wall, we have three photographs:  one of our daughter; one of our older son, and one of the two of them together.

In mathematical notation, let's call my daughter A and my older son B; so we have A, B and AB.

Now we have three children, and I have started considering how many photos we would need to show the same range of variations... and it's more than I thought.  Let's introduce the baby as C.

We would have:
A, B, C - each child individually
AB, AC, BC - each child with one other sibling
ABC - all three children together.

So the total number of pictures has gone from 3 to 7.

Let's suppose we have four children, A, B, C and D, and we want the same range of photos, with all variations.  The list grows dramatically:

A, B, C and D - individual photos - 4
AB, AC, AD, BC, BD, CD - pairs - 6
ABC, ABD, ACD, BCD - trios - 4
ABCD - group - 1
Total = 15

Children  Photos
1   1
2   3
3   7
4   15

In order to work out the nature of the series, I looked at the differences between terms, and then the second differences (i.e. the differences between the differences).
3-1 = 2
7-3 = 4
15-7 = 8

8-4 = 4

What became clear to me at this point is that the sequence is expanding exponentially or logarithmically, and not quadratically.  And then it very quickly followed that each nth term is 2n -1 (the series 2n is immediately recognisable - 1, 2, 4, 8, 16, 32 etc).  The need to introduce the -1 suggests to me that we're excluding the photo which has no children in it.

I had not expected a logarithmic series from this starting point; in fact I had not expected expected the number of photographs to increase so quickly - the volume more than doubles each time, as we have to account for every combination incorporating the new child. I was expecting something similar to a Fibonacci series - but that's more about multiplying rabbits, not children!

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