Simple as that

Consider three natural numbers \(n_{1} ,n_{2}, n_{3}\) defined as

\(- n_{1`}=x_{1}x_{2}x_{3}x_{4}\) ( \(4~digit~number\))

\(- n_{2}=y_{1}y_{2}y_{3}\) (\(3~digit~number\))

\(- n_{3}=z_{1}z_{2}\) (\(2~digit~number\))

where \(x_{1}....x_{4},y_{1}...y_{3},z_{1},z_{2}\) are all non zero digits.

Then total number of nine digit numbers formed by picking up the digits used in the numbers \(n_{1},n_{2},n_{3}\) if digits are taken only from the right side of either of the numbers \(n_{1},n_{2}~and~n_{3}\)


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