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Russian Cubes

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N000ughty Thoughts

Factorial one hundred (written 100!) has 24 noughts when written in full and that 1000! has 249 noughts? Convince yourself that the above is true. Perhaps your methodology will help you find the number of noughts in 10 000! and 100 000! or even 1 000 000!

Counting Binary Ops

Stage: 4 Challenge Level: Challenge Level:2 Challenge Level:2

Another superb solution from Andrei, School: 205, Bucharest, Romania.

For four terms, I found the following possible combinations:
(a (b (c d)))
(a (b (c d)))
(a ((b c) d))
((a b) (c d))
((a (b c))d)
and (((a b) c) d) i.e. five.

For five terms, there are 14 combinations:
(a(b(c(de)))) (a(b((cd)e)))
(a (b (c (d e)))) (a (b ((c d) e)))
(a ((b c) (d e))) (a ((b (c d)) e))
(a (((b c) d) e)) ((a b) (c (d e)))
((a b) ((c d) e)) ((a (b c)) (d e))
((a (b (c d))) e) ((a ((b c) d)) e)
((a b) c) (d e)) (((a b) (c d)) e)
(((a (b c)) d) e) ((((a b) c) d) e)

For six terms, there are 42 combinations.

I found that these are Catalan numbers, that appear from a great variety of mathematical problems, as the number of ways a regular polygon with n sides could be divided into (n-2) triangles, the number of paths of length 2n through an n by n grid that do not rise above the main diagonal, and, of course the number of ways numbers in a sequence could be grouped using binary operations, as is the case of this problem.

The first terms of the sequence of Catalan numbers are:
1, 2, 5, 14, 42, 132, 429, 1430, 4862, ...

They could be computed using the formula: $$Cat_n={^{2n}C_n \over (n+1)}= {(2n)!\over (n+1)!n!}$$

The correspondence with the problem is that n here must be the number of binary operations, i.e. the number of terms minus one.