If we can find a way to break X up as \(X = X_{1} \cup X_{2} \cup \cdots \cup X_{n}\), where each Xi is easier to count than X, then the addition principle gives an answer of \(|X| = |X_{1}|+|X_{2}|+|...If we can find a way to break X up as \(X = X_{1} \cup X_{2} \cup \cdots \cup X_{n}\), where each Xi is easier to count than X, then the addition principle gives an answer of \(|X| = |X_{1}|+|X_{2}|+|X_{3}|+ \cdots +|X_{n}|\). The answer will be |X|, so our task is to find |X|. Put \(X = X_{1} \cup X_{2} \cup X_{3} \cup X_{4} \cup X_{5}\),where \(X_{i}\) is the set of those numbers in X whose ith digit is 6, as diagramed below.
