Irregular polyhedra

[Philip Davies]

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Tomorrow, we have a Winter Fair at church and I am giving some of these to sell. I found the idea on Pinterest and I thought it a good way to use up offcuts, particularly arttractive woods. They were posted as “tsumi-ishi” and can be bought on Etsy from Japan for a surprising amount of dollars.

I was thinking of what to say to the vendor. Customers are likely to want to experiment with stacking them, that’s the idea. But they will have to be tactfully discouraged. Then I realised I had unfortunately forgotten how to calculate the theoretical number of permutations. It is certain that someone on HMT knows.

Assuming that each block has seven faces and a customer is going to emulate one of the columns of 5.
7! ,then, if the order is predetermined. Is it 7! To the power of 5, if any order is allowed?

The stack in the foreground is included, with the base wrapped in masking tape, since the best way of gluing, I found, was “Mitre Mate” , a 5-second setting superglue. But the catalyst has a tendency to wick beneath the paper, marring the appearance.

[mklotz]

Assumptions: 7 blocks in set, 5 blocks selected and that subset used to build stack

The number of combinations of n things taken m at a time is given by n! / [(n-m)!*m!]

so number of unique subsets = 7! / [2!*5!] = 6*7 / 2 = 21

with each subset, we have 5 choices for the first block, 4 for the second,etc.... so 5! = 120 arrangements.

Therefore, given the initial assumptions, 21 * 120 = 2520 unique stacks

[Philip Davies]

Thanks. I knew that I could rely upon you. However, the first block, of 5, has minimum 7 facets, each of which could theoretically be the base. (It was a mistake to say that the stacking order was predetermined) Each of the 5 blocks could form the first base. I understand that the blocks could be arranged 120 ways. But each has 7 facets, any one of which could be the first to form the base. Pardon me, since I don’t remember the maths I learned 50 years ago, and also been on my feet for 13 hours. initially I thought that would be 7! X 5! Have I missed something?

[mklotz]

Thanks. I knew that I could rely upon you. However, the first block, of 5, has minimum 7 facets, each of which could theoretically be the base. (It was a mistake to say that the stacking order was predetermined) Each of the 5 blocks could form the first base. I understand that the blocks could be arranged 120 ways. But each has 7 facets, any one of which could be the first to form the base. Pardon me, since I don’t remember the maths I learned 50 years ago, and also been on my feet for 13 hours. initially I thought that would be 7! X 5! Have I missed something?

If each of the 5 blocks can be oriented in 7 different ways, then it seems to me that would add a factor of 7^5 = 16807 to the result for each stack.

2520 * 16807 = 42,353,640

A staggering number of unique stacks, but the average player is unlikely to pay much attention to block orientation.

That 42 million number is only good for advertising copy lies. Many of the blocks will have facets unusable as bases because the center of mass is outside the footprint, i.e., they can't stand on that surface.