Particle man

I found the following problem at http://www.artofproblemsolving.com/Store/products/intro-counting/posttest.pdf and I think it's great:

Particle Man is at the origin in three-dimensional space. In how many ways can Particle Man take a series of 12 unit-length steps, each step parallel to one of the coordinate axes, from the origin to (3, 4, 5) without passing through the point (2, 3, 2)?

Slightly harder: what if Particle Man also is not allowed to pass through the point (1,1,1)?

Note by Tim Vermeulen
7 years, 11 months ago

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4 votes

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Comments

At first consider the total number of ways Particle Man can take a series of 12 unit-length steps from the origin to (3,4,5) (3, 4, 5) , i.e remove the constraint first. Consider a permutation of 3 3 identical x x s, 4 4 identical y y s, and 5 5 identical z z s. For example, xxyyyyxzzzzz xxyyyyxzzzzz is such a permutation. In each such permutation, a x x corresponds to moving one step along x x axis, a y y corresponds to moving one step along y y axis, and a z z corresponds to moving one step along z z axis. We are talking about moving positively along the axes here, because if a negative step is taken then the total number of steps taken to reach the destination will be more than 12 12 . The total number of permutations is 12!3!4!5! \frac{12!}{3!4!5!} . Now we find the number of ways Particle Man can reach his destination passing through (2,3,2) (2, 3, 2) . Consider the first part of his journey, i.e his journey from origin to (2,3,2) (2, 3, 2) . In a similar argument, we can prove that Particle Man can go there in 7!(2!)23! \frac{7!}{(2!)^23!} ways. Now consider the second part of his journey, i.e his journey from (2,3,2) (2, 3, 2) to the destination. Similarly we can prove that this can be done in ((3−2)+(4−3)+(5−2))!(3−2)!(4−3)!(5−2)!=5!3! \frac{ ((3-2) + (4-3) + (5-2))!}{(3-2)! (4-3)! (5-2)!} = \frac{5!}{3!} ways. Hence the total number of ways Particle Man can reach his destination via (2,3,2) (2, 3, 2) is 7!5!(2!)2(3!)2 \frac{7!5!}{(2!)^2(3!)^2} (the numbers are multiplied). We have to subtract this from the total number of ways, so the final answer will be 12!3!4!5!−7!5!(2!)2(3!)2 \frac{12!}{3!4!5!} - \frac{7!5!}{(2!)^2(3!)^2} .

I will post my answer to the harder part shortly.

Sreejato Bhattacharya - 7 years, 11 months ago

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That's a wonderful answer, and it is correct, obviously. On Project Euler (a website with programming challenges) a similar problem was posted, but it is much, much harder: it involves many more points in the grid that have to do with pythagorean triples. :) Here it is: http://projecteuler.net/problem=408

Tim Vermeulen - 7 years, 11 months ago
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