{Combinatorics} Coins distributed among people.

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SUMMARY

The discussion focuses on the combinatorial problem of distributing 55 identical coins among three individuals, ensuring each person receives an odd number of coins. The solution involves transforming the problem into finding the number of ways to distribute 29 units (after accounting for the odd distribution) using the Stars and Bars theorem. The correct calculation yields 378 distinct distributions, confirming the use of the formula C(29+3-1, 3-1) for this scenario.

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youngstudent16
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Homework Statement


Find the number of ways to distribute 55 identical coins among three people, so that everyone gets an odd number of coins.

Homework Equations


Stars and Bars Formula [/B]

The Attempt at a Solution



(n+r-1,n-1)

Ways to place r indistinguishable objects into n distinguishable boxes.
C(57,2)=1596 total ways [/B]


Thats about it. If it was even I could use same formula I think but with groups of objects instead. Since its odd I'm unsure.


My weak guess was I took 55/3 and used the same formula and got roughly 196 total ways I'm sure that is wrong though. Thanks for any help.
 
Last edited:
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My new attempt which came out right

My attempt thinking of it as solutions x1+x2+x3=55 solutions to the each such that each is odd z1+z2+z3=28 with no restrictions is the same as y1+y2+y3=52 such that each solution is even

Thus x1=y1+1=2z1+1

So number of ways will be 378
 
I agree with your answer but don't follow your reasoning !
Mine is that if you've got ##p_1,p_2,p_3## coins in each box, all these numbers being odd, you can write ## p_i = 2q_i - 1, \ q_i \ge 1 ##. Therefore your problem is equivalent to ## q_1 + q_2 + q_3 = 29 ## which is a classic situation of 'stars and bars' theorem.
 
youngstudent16 said:
z1+z2+z3=28
You mean =26, right? The 2 gets added later.
 

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