Maths - Help!

[BLCalliente]

We have movie night every week and there are 12 attendees. Sometimes they don't all attend. We have 12 different seats.

Just for kicks I thought it would be fun to figure out how many different seating combinations there are, including when some people don't show up. So we'd have between 0-11 empty seats on any given night, and any of the 1-12 individuals could be sitting in any different seat.

One (or more) of you mathy individuals can probably come up with the answer to this in 2 seconds. I thought it might be something like 12 factorial, but I'm rusty at the maths these days.

Challenge accepted?

[Savagious]

We have movie night every week and there are 12 attendees. Sometimes they don't all attend. We have 12 different seats.

Just for kicks I thought it would be fun to figure out how many different seating combinations there are, including when some people don't show up. So we'd have between 0-11 empty seats on any given night, and any of the 1-12 individuals could be sitting in any different seat.

One (or more) of you mathy individuals can probably come up with the answer to this in 2 seconds. I thought it might be something like 12 factorial, but I'm rusty at the maths these days.

Challenge accepted?

that is one ridiculously large number...

[Tirivaria]

12P12 + 12P11 + 12P10 + ... + 12P1 + 12P0

Where nPx = n!/(n-x)!

Edit: that last term should probably be 1 instead of 12P0. We don't really care how many different ways empty chairs can be arranged if no one shows up

[Twoddle]

It's a sum of permutations as far as I can see.

The formula for k permutations of n is n! / (n-k)!

When 1 person turns up there are 12P1 = 12 possible arrangements.
When 2 people turn up there are 12P2 = 132 possible arrangements.
When 3 people turn up there are 12P3 = 1320 possible arrangements.
When 4 people turn up there are 12P4 = 11880 possible arrangements.
When 5 people turn up there are 12P5 = 95040 possible arrangements.
When 6 people turn up there are 12P6 = 665280 possible arrangements.
When 7 people turn up there are 12P7 = 3991680 possible arrangements.
When 8 people turn up there are 12P8 = 19958400 possible arrangements.
When 9 people turn up there are 12P9 = 79833600 possible arrangements.
When 10 people turn up there are 12P10 = 239500800 possible arrangements.
When 11 people turn up there are 12P11 = 479001600 possible arrangements.
When 12 people turn up there are 12P12 = 479001600 possible arrangements.

The sum of all those is 1302061344 and that's your answer unless I've made a mistake.

[Tirivaria]

I think Twoddle and I both made a conceptual mistake.

Take the case where 3 people show up. 12P3 give the number of ways you can select 3 from 12 where order matters. It's like the ways gold, silver, and bronze can be awarded among 12 competitors. But in this case what we actually need is the number of different 3 person groups out of 12 (nCx) multiplied by the number of ways those three people can occupy 12 different chairs (12P3 gives the number of ways 3 out of twelve people can occupy 3 chairs).

[BLCalliente]

This seems like it's going to be a VERY large number.

EDIT: Thank you for giving it a whirl!

[Twoddle]

I think Twoddle and I both made a conceptual mistake.

Take the case where 3 people show up. 12P3 give the number of ways you can select 3 from 12 where order matters. It's like the ways gold, silver, and bronze can be awarded among 12 competitors. But in this case what we actually need is the number of different 3 person groups out of 12 (nCx) multiplied by the number of ways those three people can occupy 12 different chairs (12P3 gives the number of ways 3 out of twelve people can occupy 3 chairs).

Yup, the formula for k combinations of n is n! / (k!(n-k)!)

It seems you have to take this sum then:

12C1 * 12P1 +
12C2 * 12P2 +
12C3 * 12P3 +
12C4 * 12P4 +
12C5 * 12P5 +
12C6 * 12P6 +
12C7 * 12P7 +
12C8 * 12P8 +
12C9 * 12P9 +
12C10 * 12P10 +
12C11 * 12P11 +
12C12 * 12P12

Someone else can do it .

[BLCalliente]

Your mathematical formula notation is beyond my aged brain.

[Twoddle]

Is just shorthand. 12C3 for example stands for the number of ways 3 people can turn up out of the 12. 12P3 stands for the number of ways 3 people can sit themselves in a room with 12 seats. 12P3 is higher than 12C3 because the order is taken into account.

[BLCalliente]

Excellent! Thank you very much!

[time0ut]

Sounds like homework.

[BLCalliente]

Sounds like homework.

Not homework.