In a certain examination \(6\) papers are set , and to each are assigned \(100\) marks as a maximum. The number of ways in which a candidate may obtain \(40\%\) of the whole marks is \[\frac { 1! }{ A! } \left\{ \frac { (24B)! }{ (24C)! } -6.\frac { (D44)! }{ (13E)! } +15.\frac { (4F)! }{ (3G)! } \right\} \] Find the value of \[A+B+C+D+E+F+G\]

**Details and Assumptions :**

\(A,B,C,D,E,F,G\) are digits from \(1\) to \(9\)

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