## Christmas Presents

moorfield
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### Christmas Presents

I think this one does the rounds in most offices at this time of year, it is in ours today, so here's my appropriated variation of it ...

Under the tree this Christmas the three moorfield juniors will have twelve presents each, of different sizes, to open.

Usually there is a flurry of them all being opened at the same time, but this year we've decided to be a little more orderly and let them open one,
more than one, or all of their presents in turn, provided that they start from the smallest up to the largest.

How many different ways are there of opening all the presents?

Gengulphus
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### Re: Christmas Presents

moorfield wrote:Under the tree this Christmas the three moorfield juniors will have twelve presents each, of different sizes, to open.

Usually there is a flurry of them all being opened at the same time, but this year we've decided to be a little more orderly and let them open one,
more than one, or all of their presents in turn, provided that they start from the smallest up to the largest.

How many different ways are there of opening all the presents?

Let me pose an extra puzzle: how many different ways are there to reasonably interpret that puzzle?

After looking at just its first paragraph, I can see at least two different ways to interpret it that both look reasonable to me, and arguably a third. If I call the sizes of the first moorfield junior A1, A2, A3, ... A12, those of the second B1, B2, B3, ... B12 and those of the third C1, C2, C3, ... C12, they are:

1) The moorfield juniors have twelve presents each, and the sizes of the 36 presents are all different - i.e. all of the Ais, Bis and Cis are different from each other.

2) The moorfield juniors have twelve presents each, and the sizes of each junior's twelve presents are all different from each other - i.e. all of the Ais are different from each other, all of the Bis are different from each other, and all of the Cis are different from each other, but it's possible that an Ai is the same as a Bi, an Ai is the same as a Ci, or a Bi is the same as a Ci.

3) The moorfield juniors have twelve presents each, and there are twelve different sizes of present, with each junior having one present of each size - i.e. all of the Ais are different from each other, and the collection of all the Bis and the collection of all the Cis are each a rearrangement of the collection of all the Ais. I don't think this is as natural an interpretation, but I'm not entirely convinced it can be ruled out...

Moving on, the second paragraph doesn't seem clear about whether the smallest->largest condition applies to each junior independently, so that it's OK for junior A to open a present of size X before junior B has opened a present of size Y < X, or whether it applies to them all together, so that the same is not OK. And there are various other somewhat unclear points, such as who chooses how many presents are to be opened and when they make that choice: I think the most natural interpretation is that each junior chooses the number they will open each time their turn comes around, but there are certainly others, e.g. that moorfield says at the start of each round "This round, I'll let you each open only one present" or "This round, I'll let you each open more than one present" or "This round, I'll let you each open all your presents".

There are quite a few combinations of those (and other) interpretations of various aspects of the question, and they lead to quite a few different answers - more than I really care to take the time to post!

Gengulphus

jfgw
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### Re: Christmas Presents

While there is some ambiguity, I suggest that there is one interpretation which is much more obvious as a puzzle than the others. I shall assume that, while each junior has to open his or her own presents in size order, there is no size-related restriction upon who unwraps which present when.

There are 36 "unwrapping events". Junior A performs 12 of these events. There are (36!/24!)/12! ways in which these 12 events can be chosen from the 36 where only one order is permitted.

Junior B performs another 12 of these events. There are (24!/12!)/12! ways in which these 12 events can be chosen from the remaining 24 where only one order is permitted.

There is only 1 way to assign the remaining 12 events to Junior C.

Multiply these together and we get [(36!/24!)/12!]*[(24!/12!)/12!]*1
=36!/24!*24!/12!/12!/12!
=36!/12!/12!/12!
=3384731762521200

There are 3384731762521200 different ways of opening the presents.

Julian F. G. W.

Stonge
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### Re: Christmas Presents

One .

scotia
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### Re: Christmas Presents

I probably have misunderstood this puzzle - but surely it is 36.
I wait to be flagellated by the mathematicians.

jfgw
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### Re: Christmas Presents

scotia wrote:I probably have misunderstood this puzzle - but surely it is 36.
I wait to be flagellated by the mathematicians.

It will be a lot more than 36.

Consider a simpler problem:
The three juniors, A, B and C, each take the smallest one of their presents. They then unwrap them one at a time. There are six ways of doing this:
ABC,
BCA,
CAB,
CBA,
BAC,
ACB.
They repeat this with the second-smallest presents. There are, similarly, six ways of doing this. Already, after just two presents each, there are 6*6=36 ways in which they could be unwrapped.

With 12 presents each, there are 6^12 = 2176782336 ways.

The original question allows each junior to unwrap multiple presents without any other juniors opening any of theirs so the number of ways is a lot greater.

Julian F. G. W.

scotia
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### Re: Christmas Presents

jfgw wrote:
scotia wrote:I probably have misunderstood this puzzle - but surely it is 36.
I wait to be flagellated by the mathematicians.

It will be a lot more than 36.
Consider a simpler problem:
The three juniors, A, B and C, each take the smallest one of their presents. They then unwrap them one at a time. There are six ways of doing this:
ABC,
BCA,
CAB,
CBA,
BAC,
ACB.
They repeat this with the second-smallest presents. There are, similarly, six ways of doing this. Already, after just two presents each, there are 6*6=36 ways in which they could be unwrapped.
With 12 presents each, there are 6^12 = 2176782336 ways.
The original question allows each junior to unwrap multiple presents without any other juniors opening any of theirs so the number of ways is a lot greater.
Julian F. G. W.

OK - I messed up the start position - there are 6 start positions for each method (not 3 as I proposed).
Lets start with 1 present each - 6 start positions, but once this has been selected, surely the order of each is fixed - so we have 6 possible orders.
Now start with 2 presents each - again the order is fixed.
Continue up to 12 - so getting 72 (not 36 as I originally proposed).
However this proposes that all 36 presents differ in size - otherwise the number of different orders will be reduced. To take the limiting case, assume that all three persons are given identical sets, then I think this reduces the overall number to 12.
Or am I barking up the wrong tree?

jfgw
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### Re: Christmas Presents

The way I interpreted the question, it is only the order in which each junior opens his or her presents that is fixed.

The sequence could be, for example,
ABCAAAACBCBABBCCCAAAABBCCCBAABBCBBCC.

The juniors could be more orderly and open their presents thus:
ABCABCABCABCABCABCABCABCABCABCABCABC.

Or the bigger juniors could bully the smaller ones and the order could be:
AAAAAAAAAAAABBBBBBBBBBBBCCCCCCCCCCCC.

Julian F. G. W.

scotia
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### Re: Christmas Presents

jfgw wrote:The way I interpreted the question, it is only the order in which each junior opens his or her presents that is fixed.
The sequence could be, for example,
ABCAAAACBCBABBCCCAAAABBCCCBAABBCBBCC.
The juniors could be more orderly and open their presents thus:
Julian F. G. W.

I was not allowing the number of openings to change once the action has started.
So I am happy with ABCABCABCABCABCABCABCABCABCABCABCABC.
Or AAAAAAAAAAAABBBBBBBBBBBBCCCCCCCCCCCC, but not ABCAAAACBCBABBCCCAAAABBCCCBAABBCBBCC.
Clearly the rules require to be specified more tightly - as proposed by Gengulphus

IrishIceHawk
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### Re: Christmas Presents

moorfield wrote:I

Under the tree this Christmas the three moorfield juniors will have twelve presents each, of different sizes, to open.

Usually there is a flurry of them all being opened at the same time, but this year we've decided to be a little more orderly and let them open one,
more than one, or all of their presents in turn, provided that they start from the smallest up to the largest.

How many different ways are there of opening all the presents?

The way I interpreted this was that each Junior had 12 presents of different size, and the size order only apples to the particular Junior.
i.e. it's a way of saying "who want's to open a present next? (random), okay, you can open one of yours, so long as it's the smaller of yours)

That in mind, I see it as:
number of total possible openings, without order is 36!
However, this include invalid orders for each junior,
Take an example of junior A opening all presents first, so opens presents in position 1-12
There will be 12! options of A1..A12, followed by the rest of the openings

However, only one of those will be in the correct order, so only 1/12! of these possibilities is valid, and the same goes if the juniour opens in any other positions.
Applying the same for the other juniors, it reduces the valid answers by the same ratio.

So the total valid possibilities is 36! / (12! * 12! * 12!) which I work out as:
479,001,600

Regards,
Martin

moorfield
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### Re: Christmas Presents

IrishIceHawk wrote:So the total valid possibilities is 36! / (12! * 12! * 12!) which I work out as:
479,001,600

Apologies I'd completely forgotten to return to this thread. But yes it is essentially about merging ordered lists - agree badly worded. The formula I've seen for two lists size n,m is (n+m)!/(n!m!), and that's the answer I came up with too.

jfgw
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### Re: Christmas Presents

IrishIceHawk wrote:So the total valid possibilities is 36! / (12! * 12! * 12!) which I work out as:
479,001,600

I had essentailly the same formula and came to a different answer,

jfgw wrote:Multiply these together and we get [(36!/24!)/12!]*[(24!/12!)/12!]*1
=36!/24!*24!/12!/12!/12!
=36!/12!/12!/12!
=3384731762521200

479,001,600 is actually 12! so probably a simple error.

36!=371,993,326,789,901,217,467,999,448,150,835,200,000,000

12!=479,001,600

(12!)^3=109,903,340,320,478,724,096,000,000

36!/(12!^3)
=371,993,326,789,901,217,467,999,448,150,835,200,000,000/109,903,340,320,478,724,096,000,000
=3,384,731,762,521,200

Julian F. G. W.