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Could anyone help me understand this argument?

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Hi,
This problem is driving me crazy. All this wrong solution is doing is confuse me. What does the author mean by overflow toy? :s-smilie:
How are we over-counting by a factor of 3? :confused:

How do we still end up counting a choice where Fred gets the puzzle, Bill gets the doll, and Joe gets the video game as a different choice! :unsure:

Thank you!!

Edit:
I do not want a solution to the problem. I can solve this problem considering 2 mutually exclusive cases, namely (2 boys, 1 girl) and (3 boys).
I am trying to understand why the particular method (seemingly right) is wrong. Any other explanation showing that the method does not work would be appreciated.
(edited 5 years ago)
Reply 1
Is the answer 6^3? If so, its fairly straight forward (couple of lines).
But yeah, the discussion does obfuscate things :-)

Edit missed the "at least" from the question.
(edited 5 years ago)
Original post by mqb2766
Is the answer 6^3? If so, its fairly straight forward (couple of lines).
But yeah, the discussion does obfuscate things :-)

The numerical answer is 240. Any idea on what the author is talking about?
(edited 5 years ago)
Original post by Quantum Horizon
The numerical answer is 240. Any idea on what the author is talking about?


The three toys are being allocated in a particular order. The first two toys are allocated to the boys. The "overflow toy" is the name they're giving to the third toy allocated, and it could go to either a boy or a girl.

If a different order had been chosen we would have a different leftover toy. Hence the x3.

If the leftover toy goes to a girl - no problem.

IF it goes to a boy, we're triple counting. Since:

If the order of toys was "video, doll, puzzle", and we allocate "Joe, Bill, Fred", then this is the same as the order of the toys being, for example, "puzzle, doll, video", and allocating to "Fred, Bill, Joe".
Reply 4
Original post by Quantum Horizon
The numerical answer is 240. Any idea on what the author is talking about?


1/2 an answer and its probably better to split it up into the two cases BBB, BBG as this is the main cause of the confusion?

BBB
Using the authors reasoning, there are
4*3*2*3 different combinations of presents / boys
This is wrong, there should just be
4*3*2
Using the 4, 3, 2 argument of 4 boys, 3 boys, 2 boys, you've got to divide by 3! to make the boy order unimportant (4 different groups). Then multiply by 3! to make the present / person combination correct. So there are 4*3*2 different ways of giving out the presents to the 3 boys.

This is the overcounting the author refers to. As far as I can see it, its just simply a case of order important/unimportant and handling it correctly. The same divisor does not apply to the BBB & BBG subgroups which is the "other" mistake.

BBG similar reasoning about the person order and present combinations. You have 4*3 (order important) combinations of 2 boys or 6 order unimportant boys. Multiplied by 6 girls and then 3! presents gives the numbers correct.
Original post by ghostwalker

If the order of toys was "video, doll, puzzle", and we allocate "Joe, Bill, Fred", then this is the same as the order of the toys being, for example, "puzzle, doll, video", and allocating to "Fred, Bill, Joe".


PRSOM!
You're amazing.
Original post by mqb2766
1/2 an answer and its probably better to split it up into the two cases BBB, BBG as this is the main cause of the confusion?

BBB
Using the authors reasoning, there are
4*3*2*3 different combinations of presents / boys
This is wrong, there should just be
4*3*2
Using the 4, 3, 2 argument of 4 boys, 3 boys, 2 boys, you've got to divide by 3! to make the boy order unimportant (4 different groups). Then multiply by 3! to make the present / person combination correct. So there are 4*3*2 different ways of giving out the presents to the 3 boys.

This is the overcounting the author refers to. As far as I can see it, its just simply a case of order important/unimportant and handling it correctly. The same divisor does not apply to the BBB & BBG subgroups which is the "other" mistake.

BBG similar reasoning about the person order and present combinations. You have 4*3 (order important) combinations of 2 boys or 6 order unimportant boys. Multiplied by 6 girls and then 3! presents gives the numbers correct.

Thank you so much!! It makes sense now.
Really appreciate it!!!

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