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In the name of Allah
In how many ways can you choose three numbers from 1 to 100 with different sum, If you can choose a number more than one time?
Eg: {3,2,1} is same as {1,1,4} because 1+1+4=3+2+1.
I know how to calculate how many ways is there to choose three numbers if you can choose a number more than one time by counting the answers of the equation:
x1 + x2 + x3+ x4 +...x100 = 3 (If xi is a whole number.)
But I have no idea about this one.

madfd adfd
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  • You can obtain all sums from $3$ to $300$. – peterwhy Oct 10 '23 at 18:26
  • It is highly unusual to consider ${3,2,1}$ the same as ${1,1,4}$... but if that is really what you want... you are asking how many terms would be in the expansion of $(x+x^2+x^3+\dots+x^{100})^3$ after simplification. It should be clear that every result from $3$ on up to $300$ is possible. – JMoravitz Oct 10 '23 at 18:26
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    If you need further convincing... consider the sequence of outcomes: $(1,1,1),(1,1,2),(1,1,3),(1,1,4),\dots,(1,1,99),(1,1,100),(1,2,100),(1,3,100),(1,4,100),\dots,(1,99,100),(1,100,100),(2,100,100),\dots,(99,100,100),(100,100,100)$ where to get the next term in the sequence we increase the last appearing element by $1$ which is able to be increased. We get a resulting sum which is exactly $1$ more than the previous sum, and this is always able to be done up until all of the numbers are $100$ at which point we can no longer continue. – JMoravitz Oct 10 '23 at 18:30

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