n-1-2n-2-3n-3-rn-r-n-n-1-n-1-n-2-n-3-n-r-1-n-1-2-n-2-3-n-3-4-n-4-r-n-r-1-0-n-1-n-2-n-3-n-Prove-the-above-identity-

Question Number 139560 by Dwaipayan Shikari last updated on 28/Apr/21

\underset{n_{1} + 2 n_{2} + 3 n_{3} + . . + {r n}_{r} = n}{\sum^{n}} \frac{1}{n_{1}! n_{2}! n_{3}! . . n_{r}! 1^{n_{1}} 2^{n_{2}} 3^{n_{3}} 4^{n_{4}} \dots r^{n_{r}}} = 1

0 ⩾ n_{1}, n_{2}, n_{3}, . . ⩾ n

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