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Today Puzzle #716

Puzzle No. 716鈥� Monday 13 April

Yesterday we had an egg hunt and I hid ten Easter eggs in piles along the path in my garden. I could have hidden them singly, or in pairs or in piles of any size. And the individual piles could have been in any order along the path. There were lots of possibilities. How many exactly?

Today’s #PuzzleForToday has been set by Hugh Hunt, Reader in Engineering Dynamics and Vibration at Trinity College, Cambridge

Click here for the answer

512

There's a pattern and the number of arrangements goes as 2^(N-1) for N eggs:

one egg (1 = 2^0)

1

two eggs (2 = 2^1)

2

11

three eggs (4 = 2^2)

3

12 21

111

four eggs (8 = 2^3)

4

13 31 22

112 121 211

1111

five eggs (16 = 2^4)

14 41 23 32

113 131 311 122 212 221

1112 1121 1211 2111

11111

It seems that for N eggs there are 2^(N-1) combinations

This is not hard to prove.

so for N=10 the answer is 512

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