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What is number of ways in which 6 pencils can be distributed between two boys if pencils are indistiguishible?

I think that we have to divide 6 pencils in two groups, so we have to do $\binom66+\dots+\binom66=2^6$, but that is the number of ways when they are distinguishible!

Silent
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    If pencils are indistinguishable, then it only matters how many one boy got. Right? He can get 0,1,2,3,4,5,6 (then the other one gets 6,5,4,3,2,1,0). – Poppy Jan 21 '14 at 05:33
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    This a simple version (x+y=6 - number of integer solutions) of the following problem: how many integer solutions are there for equation $x_1+...+x_m=k$ without any restrictions? http://math.stackexchange.com/questions/229430/how-many-integer-solutions-to-a-linear-combination-with-restrictions – Poppy Jan 21 '14 at 05:49

2 Answers2

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The answer is $7$: if you give $x$ pencils to the first boy you must give $6-x$ to the second boy. As $x$ is a number between $0$ and $6$, there are $7$ possibilities.

user642796
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Zur Luria
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There are only $7$ patterns as $$(\text{the first boy, the second boy})=(6,0),(5,1),(4,2),(3,3),(2,4),(1,5),(0,6).$$

In general, if pencils are indistiguishable, then the number of ways in which $n$ pencils can be distributed to two boys is $n+1$.

mathlove
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  • Thank you so much, Sir, just one more question, it may be pathetic, but I can't understand: My book says," unordered sampling with replacement: The general formula for the number of unordered samples of size k with replacement from n elements is $\binom{n+k-1}{n}$" Why is not this problem unordered sampling with replacement? – Silent Jan 21 '14 at 06:00
  • Your formula is about number of combinations with repetition, right? see here. http://en.wikipedia.org/wiki/Combination#Number_of_combinations_with_repetition Maybe you wanted to write $\binom{n+k-1}{k}$, right? Then, you can use it as $k=6,n=2.$ – mathlove Jan 21 '14 at 06:09
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    Really sorry, sir! it is as yours! – Silent Jan 21 '14 at 06:12