Finding the least number of plants watered by three gardeners

Question

There are 100 plants and three gardeners A, B, and C. If A watered 68 plants, B watered 78 plants, and C watered 88 plants, at least how many plants are watered by all three gardeners?

I drew a Venn diagram and let $|A \cap B| - |A \cap B \cap C| = |A \cap C| - |A \cap B \cap C| = 34.$ Then, I let $|B \cap C| - |A \cap B \cap C| = 44,$ so that meant that $|C| - |A \cap C| - |B \cap C| = 10.$ However, this totaled to $122,$ so I concluded that the answer was $122 - 100 = 22.$ Can somebody tell me if my logic is correct or not? Thanks.

Answer 1

Hint: 32 plants weren't watered by A, 22 plants weren't watered by B, and 12 plants weren't watered by C. The number of plants watered by all three gardeners will be minimized when there is no overlap between those three sets of plants.

Let $A'$, $B'$, and $C'$ represent the sets of plants not watered by A, B, and C respectively. The set of plants watered by all three is clearly the complement of $A'\cup B'\cup C'$. Thus, to minimize the size of that set, we must strive to maximize the size of $A'\cup B'\cup C'$. Since the sizes of $A'$, $B'$, and $C'$ are fixed, their union has maximum size when the sets are disjoint. Therefore, the minimum number of plants watered by all three gardeners is $100-(12+22+32)=34$.