# 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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