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In how many ways we can distribute 101 coins to three friends such that sum of the coins of two friends is more than or equal to the number of coins of third friend. my views:should I distribute 50 and then 51 ,52 coins ....is there any elegant way please guide

user344005
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    Do you mean the sum of any two friends is more than the third or specific friends: $friend_1$ and $friend_2$ together are more than or equal to $friend_3$? – Ian Miller Jun 02 '16 at 03:02
  • This has something to do with the triangle inequality... – Jared Jun 02 '16 at 03:15
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    Additional query - can a friend receive zero coins or only strictly positive amounts? – Ian Miller Jun 02 '16 at 03:26
  • Ian, no. Because if so, then one of the other friends have more coins than the other. Adding zero and the smallee of the two numbers is less than the greater number, which contradicts what we want to count. – Burrrrb Jun 02 '16 at 03:41
  • @Asemismaiel I was hoping for an comment addressing both queries. If its any two friends then ${0,50,51}$ does get counted (if zero is allowed) as $0+51>50$. – Ian Miller Jun 02 '16 at 04:26
  • @IanMiller no is not correct because $ 0 + 50 < 51 $ – Anonymous Jun 02 '16 at 04:51
  • @Anonymous Hence why I asked if it is 'any two friend' or 'specific friends'. If he just wants any two friends to have more than the third person it is different than if he wants friends A and B to have more than friend C which is different again to all combinations of 'any two friends' more than the other. – Ian Miller Jun 02 '16 at 04:55
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    @IanMiller I believe since the problem doesn't mention any labels for friends the "third" or "ordered" friend is arbitrary. – Anonymous Jun 02 '16 at 04:58
  • This is http://oeis.org/A069981, where it says $a(n)=(n+8)(n-2)/8$ for $n$ even and $(n^2-1)/8$ for $n$ odd. Plugging in $n=101$ gives $a(n)=1275$ – Ross Millikan Jun 02 '16 at 05:22

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$$ x+y+z=101 $$ $$ x+y \geq z \quad x+z \geq y \quad y+z \geq x $$ $$ x+y= 101-z \geq z \rightarrow z\leq 50 $$ $$ x+z= 101-y \geq y \rightarrow y\leq 50 $$ $$ y+z= 101-x \geq x \rightarrow x\leq 50 $$

$$ x = 50 - u \quad y = 50 - v \quad z = 50 - w $$ $$ (50 - u) + (50 - v) + (50 - w) = 101 $$ $$ u + v + w = 49 \Rightarrow 0 \leq u + v = 49 - w \leq 49 $$ $$ u+v = i : \{0\leq i \leq 49\} $$

$$ u+v = i \rightarrow (u = 0, v = i) \dots (u=i, v=0) \Rightarrow i+1 \mbox{ways} $$

$$ \sum_{i=0}^{49}i+1 = 1275 $$

Anonymous
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