$12$ men and $18$ women can do a work in $10$ days. $9$ men and $18$ women can do the same work in $12$ days. If $2$ men and $3$ women together do the same work, in how many days will they complete the work?

Let the daily efficiency of $1$ man be $M$ and $1$ woman be $W$.
According to the problem, the total work is equal in both cases:

$$\text{Total Work}=(12M+18W)\times 10=(9M+18W)\times 12$$

Expand and solve the equation to find the relation between $M$ and $W$:

$$120M+180W=108M+216W$$

$$120M-108M=216W-180W$$

$$12M=36W$$

$$1M=3W$$

Now substitute $1M=3W$ to find the total work in terms of $W$:

$$\text{Total Work}=(12(3W)+18W)\times 10=(36W+18W)\times 10=54W\times 10=540W$$

Now, we need to find the time taken by $2$ men and $3$ women to complete the same work.
Combined efficiency of $2$ men and $3$ women:

$$\text{Efficiency}=2M+3W=2(3W)+3W=6W+3W=9W$$

Number of days required ($D$):

$$D=\frac{\text{Total Work}}{\text{Combined Efficiency}}=\frac{540W}{9W}=60\text{ days}$$

Therefore, they will complete the work in $60\text{ days}$.