In a agricultural survey of 120 farmers in a community, 65 grow maize, 50 grow yam, and 42… — JAMB UTME

In a agricultural survey of 120 farmers in a community, 65 grow maize, 50 grow yam, and 42 grow cassava. Furthermore, 24 grow both maize and yam, 18 grow both maize and cassava, and 15 grow both yam and cassava. If 8 farmers grow none of these three crops, find the number of farmers who grow all three crops.

Answer: 12 farmers

Answer

12 farmers grow all three crops.

Using the 3-set inclusion-exclusion principle, the total number of farmers growing at least one crop is

120 - 8 = 112

. Expanding

|M \cup Y \cup C| = |M| + |Y| + |C| - (|M \cap Y| + |M \cap C| + |Y \cap C|) + |M \cap Y \cap C|

gives

112 = 65 + 50 + 42 - 24 - 18 - 15 + x

. Simplifying yields

112 = 100 + x

, which gives

x = 12

Step-by-Step Solution

Determine the cardinality of the union of all three sets.

|M \cup Y \cup C| = 112

Subtract the farmers who grow none of the crops from the universal set size (

120 - 8 = 112

Set up the Inclusion-Exclusion equation for three sets.

112 = 65 + 50 + 42 - (24 + 18 + 15) + x

Inclusion-exclusion states that

|A \cup B \cup C| = n(A) + n(B) + n(C) - n(A \cap B) - n(A \cap C) - n(B \cap C) + n(A \cap B \cap C)

Solve for the unknown value

x

representing the intersection of all three sets.

x = 12

Simplifying gives

112 = 100 + x

, which leads directly to

x = 12

Key Concept

Principle of Inclusion-Exclusion for 3 Sets

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