Number of onto functions ?

With $f(x_{1}) = y_{1}$ and $f(x_{2}) = y_{2}$ we must have each of $y_{3}, y_{4}$ mapped onto by at least one element of $S =$ { $x_{3}, x_{4}, x_{5}$ } for $f$ to be an onto function.

We then have two scenarios to consider:

(i) One element of $S$ maps onto either $y_{1}$ or $y_{2}$ and the other two elements are mapped onto { $y_{3}, y_{4}$ }. As there are $3$ elements in $S$ , $2$ elements in { $y_{1}, y_{2}$ } and $2$ ways in which an onto function can be mapped between two $2$ -elements sets, the number of possible onto functions in this scenario is $3*2*2 = 12$ .

(ii) All the elements of $S$ are mapped to { $y_{3}, y_{4}$ }. The are $2^{3} = 8$ ways in which the elements of $S$ can be mapped, two of which result in them all being mapped to either $y_{3}$ or $y_{4}$ . Thus there are $8 - 2 = 6$ onto functions possible in this scenario.

The total number of onto functions under the given conditions is therefore $12 + 6 = \boxed{18}$ .

The number of onto functions = ( 5 choose 2) + (4*2) =18.

P.S. Not a level 5 problem.....

since f(x1)=y1 $ f(x2)=y2,

no. of onto fun.= 3!/2! 1! multiply by 3! for arrangement.