Consider counting the number of orderings of the 26 letter English alphabet which do not include the words AMBER, BERTH, or THINK.
Let U be the universe of all orderings of the alphabet, so |U| = 26!
How can I find the number that contain AMBER, BERTH, THINK, AMBER and BERTH, AMBER and THINK, BERTH and THINK, and all three?
If a permutation contains both AMBER and BERTH, it actually contains AMBERTH. Think of this as one "block", and the other 18 letters are each their own blocks. You can permute the blocks in any order, so there are $19!$ permutations containing AMBER and BERTH. The others are similar.