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CEO
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A varaint of a MGMAT problem. A , B, and C have 5 bagels [#permalink]
 15 Dec 2003, 08:09
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A varaint of a MGMAT problem.
A , B, and C have 5 bagels to share. If any one of the them can be given any whole number of bagels from 0 to 5, in how many different ways can the bagels be distributed?
(A) 21
(B) 42
(C) 120
(D) 504
(E) 5040
Last edited by Praetorian on 15 Dec 2003, 09:06, edited 2 times in total.
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Director
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5 0 0 x3
4 1 0 x6
3 1 1 x3
3 2 0 x6
2 2 1 x3
21, I think...
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Director
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It would be great to know the way to use theory to do this problem...
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CEO
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stoolfi wrote: 5 0 0 x3
4 1 0 x6
3 1 1 x3
3 2 0 x6
2 2 1 x3
21, I think...
That is correct. Great job.
thanks
praetorian
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Director
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21, of course, is the sum of numbers from 1-6.
Which makes sense.
If person A has 0 bagels, person B can have 0,1,2,3,4,5 and person C will get whatever is left.
If person A has 1 bagels, person B can have 0,1,2,3,4 and person C will get whatever is left.
If person A has 2 bagels, person B can have 0,1,2,3 and person C will get whatever is left.
If person A has 3 bagels, person B can have 0,1,2 and person C will get whatever is left.
If person A has 4 bagels, person B can have 0,1 and person C will get whatever is left.
If person A has 5 bagels, person B can have 0 and person C will get whatever is left.
But there's still gotta be a better way...
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Director
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When you say that this problem is reminiscent of an old MGMAT problem, are you referring to this thread?
http://www.gmatclub.com/phpbb/viewtopic ... ght=#17866
Because we detemined the formula for positive integers, but the addition of ZERO is the difference between that problem and this one, and I cannot figure out how to compensate/adjust for that.
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CEO
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stoolfi wrote: When you say that this problem is reminiscent of an old MGMAT problem, are you referring to this thread? http://www.gmatclub.com/phpbb/viewtopic ... ght=#17866Because we detemined the formula for positive integers, but the addition of ZERO is the difference between that problem and this one, and I cannot figure out how to compensate/adjust for that. no stoolfi, this problem is from MGMAT's archives.
let me see if i can send you the explanation.
great work
keep posting
thanks
praetorian
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Director
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Here is another approach (virtually the same as Stoolfi's approach)
With:
5 Bagels (means 0 bagels for everyone else): 3C1=3
4 Bagels (4,1,0) = 3C1*2C1=6
3 Bagels (3,2,0) = 3C1*2C1=6
3 Bagesl (3,1,1) = 3C1=3
2 Bagels (2,2,1) = 3C1=3
Sum=21
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