# Series N – New

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1225869.  Sun Feb 12, 2017 4:18 am

 Quote: If you have 17 guests and 2 tables of 10, that has 131,702 possible seating arrangements.

I don't get this. How did they work it out?

1225992.  Sun Feb 12, 2017 8:06 pm

Quote:
 Quote: If you have 17 guests and 2 tables of 10, that has 131,702 possible seating arrangements.

I don't get this. How did they work it out?

2^17?

 1226010.  Sun Feb 12, 2017 11:51 pm So why is it 2^17? It's the number of subsets of a set with 17 elements – but how does that fit with the 2 tables of 10?

 1226573.  Thu Feb 16, 2017 1:36 am The idea is that each guest can be seated in either of the two tables, and therefore every guest added should double the number of possible arrangements. (Or, if you'd rather look at it this way: if you have a set of 17 guests, the people who are seated at table A are a subset of that set, so you'll have the same number of possible seating arrangements as there are subsets in that set.) That idea is wrong, though, because it's not true that "each guest can be seated in either of the two tables" - we are told that each table can only fit ten people, so when all ten seats at one of the tables are taken, the remaining guests must be sent to the other one. Therefore many subsets are actually excluded, and the true number of seating arrangements is significantly lower than 2^17.

 1226574.  Thu Feb 16, 2017 1:47 am (Or significantly higher, if we also take into consideration the arrangement of the guests within each table.)

 1226825.  Fri Feb 17, 2017 1:25 pm Since each table can seat at most 10 guests, each must also have at least 7 guests (otherwise the other table would have to accommodate more than 10 guests); therefore the correct number of seating arrangements (ignoring the arrangement of the guests within each table) should be 17C7 + 17C8 +17C9 + 17C10 = 87,516 I just hate it whenever someone throws out some impressively big number for some calculation without explaining how they got it, and it turns out they are wrong.

 1226844.  Fri Feb 17, 2017 5:49 pm But wouldn't the seating arrangements have to also take into consideration the placement of the three blank spaces at either table - they could be either all three on table A, all three on table B, or two on A and one on B or one on A and two on B.

 1227016.  Sun Feb 19, 2017 12:37 pm Olinguito's calculation already takes that into consideration: it is based on dividing the problem into four cases according to the number of guests at table A (7, 8, 9 or 10), and each of these cases corresponds to one of these four placements. "Two blank spaces on table A and one on table B" is just another way of saying "8 guests at table A".

1227183.  Tue Feb 21, 2017 2:22 am

 Quote: (ignoring the arrangement of the guests within each table)

Hence my questionmark (2^17?). The calculation and conditions (like you not leaving when you're surrounded by toddlers) are missing, and 2^n is not the right answer indeed.

 1227400.  Tue Feb 21, 2017 6:55 pm I'd calculate this as 20x19x18x17...x5x4. This works out to several trillion possibilities. so they clearly didn't use that method. If you're determining the number of guests per table, that works out to be just 4 (7&10, 8&9, 9&8, 10&7). If we pop the numbers into an nCr formula (determines possibles combinations when taking a sample n from group r), we get a measly 1,140. I'm guessing she just pulled the number out of her @r\$e.

 1227429.  Wed Feb 22, 2017 5:23 am Is it possible that they are regarding a wedding party as consisting of specific people? That is to say, rather than just homogenous "guests" there is Aunty Joan, Cousin Bill etc, each of whom could be at either one of the tables, so that any given table of seven people isn't the only table of seven people?

1227527.  Wed Feb 22, 2017 2:53 pm

 crissdee wrote: Is it possible that they are regarding a wedding party as consisting of specific people? That is to say, rather than just homogenous "guests" there is Aunty Joan, Cousin Bill etc, each of whom could be at either one of the tables, so that any given table of seven people isn't the only table of seven people?

That would be a choice of 2, repeated 17 times. In mathematical terms it's expressed as 2^17. Which works out to 131,072. Which is close enough to the quoted number to be plausible as the source. So, I think you've nailed it.

 1227579.  Wed Feb 22, 2017 7:40 pm The discussion above already assumes the guests are specific, distinguishable people. If all the guests are identical there are only four ways (7-10, 8-9, 9-8, 10-7), rather than 87,516, to split them to the two tables.

 1229579.  Tue Mar 07, 2017 3:13 pm Just watched episode and 131072 didn't make much sense for reasons given above. I think elves looked at http://www.improbable.com/news/2012/Optimal-seating-chart.pdf which gives more context to the problem and includes the remark "There are 2^17 = 131,702 possible combinations for seating these guests." Still looks like a careless sloppy statement.

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