Mathematics - Functions Question with Solution | TestHub

Finding the cases for $f\left(a,b\right)=f\left(b,a\right)\ge a$ we have,

$\left(1,1\right),\left(1,4\right),\left(4,1\right),\left(2,4\right),\left(4,2\right),\left(3,4\right),\left(4,3\right),\left(4,4\right)$- all have one choice for image.

$\left(2,1\right),\left(1,2\right),\left(2,2\right)-$ all have three choices for image.

$\left(3,2\right),\left(2,3\right),\left(3,1\right),\left(1,3\right),\left(3,3\right)-$all have two choices for image.

So the total functions $=3\times 3\times 2\times 2\times 2=72$

Case 1: None of the pre-images have $3$ as image. 

Total functions $=2\times 2\times 1\times 1\times 1=4$

Case 2: None of the pre-images have $2$ as image

Total functions $=2\times 2\times 2\times 2\times 2=32$

Case 3: None of the pre-images have either $3$ or $2$ as image

Total functions $=1\times 1\times 1\times 1\times 1=1$

$\therefore$ Total onto functions $=72-4-32+1=37$