CAT 2024 Slot 1 · Quant · Question 7

The condition says every element of B is hit, so we are counting onto functions from a 6-element set to a 3-element set.

Without any condition, each of the 6 elements of A has 3 choices, giving {3}^{6}=729 functions.

Subtract those that miss at least one element of B, using inclusion and exclusion.

Functions missing a chosen one of the three targets use only 2 values: {2}^{6}=64 each, and there are 3 such choices, giving 3\times 64=192.

Those using only a single value have been subtracted twice, so add them back: there are 3 such functions.

The count is 729-192+3=540.

Shortcut. The standard formula for onto functions from m elements to n is \sum_{i=0}^{n}{{\left( -1 \right)}^{i}}\left( \genfrac{}{}{0pt}{}{n}{i}\right){\left( n-i \right)}^{m}. With m=6 and n=3 it reads 729-192+3 straight off, with no case analysis.