$\begin{array}{1 1} 16 \\ 36 \\ 60 \\ 180 \end{array}$

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There are 4 even places and 5 odd places.

Odd digits are 3355.

Odd numbers are placed in these 4 places in $\large\frac{4!}{2!.2!}$ ways.

Even digits are 22888.

These even digits are placed in 5 odd places in $\large\frac{5!}{3!.2!}$ ways.

$\therefore $ No. of required 9 digit numbers= $\large\frac{4!}{2!.2!}.\large\frac{5!}{3!.2!}$

$=60$

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