Let N be the number of onto functions f:\1,2,3,4,5\→\1,2,3\ satisfying f(1)≠ f(2). Find N.

Total onto functions from a 5-set to a 3-set: $3^5-3\cdot2^5+3=150$.
Subtract the onto functions with $f(1)=f(2)$. Glue $1$ and $2$ into one element: this is the number of onto maps from a 4-element set to a 3-element set, $3^4-3\cdot2^4+3=36$.
Hence $N=150-36=114$.
Alternative check: choose the ordered pair of distinct values for $f(1),f(2)$ in $6$ ways. The remaining three elements must cover the third value at least once: $3^3-2^3=19$ maps. $6\cdot19=114$. A brute-force enumeration of all $243$ functions also gives $114$.