Let S = \ 1, 2, 3, 4\ . Then the number of elements in the set f: S...

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)$- all have one choice for image.</p><p>\left(2, 1\right), \left(1, 2\right), \left(2, 2\right) -$ 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) -$all have two choices for image.</p><p>So the total functions$= 3 \times 3 \times 2 \times 2 \times 2 = 72$</p><p>Case 1: None of the pre-images have$3$as image. </p><p>Total functions$= 2 \times 2 \times 1 \times 1 \times 1 = 4$</p><p>Case 2: None of the pre-images have$2$as image</p><p>Total functions$= 2 \times 2 \times 2 \times 2 \times 2 = 32$</p><p>Case 3: None of the pre-images have either$3 or 2$as image</p><p>Total functions$= 1 \times 1 \times 1 \times 1 \times 1 = 1$</p><p>$∴$Total onto functions$= 72 - 4 - 32 + 1 = 37