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Question Number 139560 by Dwaipayan Shikari last updated on 28/Apr/21 $$\underset{{n}_{\mathrm{1}} +\mathrm{2}{n}_{\mathrm{2}} +\mathrm{3}{n}_{\mathrm{3}} +..+{rn}_{{r}} ={n}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{{n}_{\mathrm{1}} !{n}_{\mathrm{2}} !{n}_{\mathrm{3}} !..{n}_{{r}} !\mathrm{1}^{{n}_{\mathrm{1}} } \mathrm{2}^{{n}_{\mathrm{2}} } \mathrm{3}^{{n}_{\mathrm{3}}…
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