# The number of non-negative integer solutions of $a + 2b + 3c + 4d + 5e + 6f = 10000000$ [closed]

Find how many solutions are there to the equation $a + 2b + 3c + 4d + 5e + 6f = 10000000$ where $a, b, c, d, e, f$ are all non-negative integers?

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why should we find it? More seriously, what did you try? Do you have any idea where to start? Where did that question come from? –  mau Oct 2 '13 at 10:01
i have found a similar question on stackexchange The solutions of x+2y+3z=n,(x,y,z∈N) –  Agarwal Shubham Oct 2 '13 at 10:11

## closed as off-topic by Arthur, azimut, Daniel Rust, Davide Giraudo, SeiriosOct 2 '13 at 10:57

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Just over $1.1574 \times 10^{30}$
This can be solved by recursion: my Java applet would try to calculate it with "Partitions of $10000000$" with "any number of terms" and "each part no more than $6$", but seems to get vey slow on my machine for partitions of more than $4000000$.
That is the number of partitions of $n$ where each part is between 5 and 10. There will be a simple generating function to provide an answer. –  Henry Oct 2 '13 at 22:47