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A varaint of a MGMAT problem. A , B, and C have 5 bagels [#permalink]
15 Dec 2003, 07:09

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A

B

C

D

E

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct
0% (00:00) wrong based on 0 sessions

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, 08:06, edited 2 times in total.

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.

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.

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.

no stoolfi, this problem is from MGMAT's archives.

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