# find the number of solutions

$a!+b!+c! = 3^{d}$ $a<=b<=c$ a, b and c are natural numbers if the total number of quadruples of (a,b,c,d) are n, then find the sum of $a_{1} + a_{2} + ... a_{n} + b_{1} + b_{2} +.... b_{n} + c_{1} + c_{2} + .. c_{n} + d_{1} + d_{2} + ... d_{n}$

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