A group of market women sell at least one of yam, plantain and maize. 12 of them sell maize, 10 sell yam and…

Analysis

To solve this problem, we use a Venn diagram approach for three sets: Y (Yam), P (Plantain), and M (Maize). We are given the total number of people in each set and the intersections.

Concept

We need to find the total number of women, which is the union of the three sets: $n(Y \cup P \cup M)$. The formula for the union of three sets is: $n(Y \cup P \cup M) = n(Y) + n(P) + n(M) - [n(Y \cap P) + n(Y \cap M) + n(P \cap M)] + n(Y \cap P \cap M)$

Step-by-Step Solution

1. Identify the given values:

• $n(M) = 12$
• $n(Y) = 10$
• $n(P) = 14$
• $n(P \cap M) = 5$
• $n(Y \cap M) = 4$
• $n(Y \cap P \text{ only}) = 2$
• $n(Y \cap P \cap M) = 3$

2. Determine the intersections:

• Since $n(Y \cap P \text{ only}) = 2$ and $n(Y \cap P \cap M) = 3$, the total intersection $n(Y \cap P)$ is: $n(Y \cap P) = n(Y \cap P \text{ only}) + n(Y \cap P \cap M) = 2 + 3 = 5$

3. Calculate the total using the principle of inclusion-exclusion: We have all the components needed for the formula:

• Sum of individual sets: $12 + 10 + 14 = 36$
• Sum of double intersections: $n(Y \cap P) + n(Y \cap M) + n(P \cap M) = 5 + 4 + 5 = 14$
• Triple intersection: $3$

Applying the formula: Total $= (12 + 10 + 14) - (5 + 4 + 5) + 3$ Total $= 36 - 14 + 3$ Total $= 22 + 3 = 25$

Verification

If we fill the Venn diagram regions:

• All three: 3
• Yam and Plantain only: 2
• Yam and Maize only: $4 - 3 = \mathbf{1}$
• Plantain and Maize only: $5 - 3 = \mathbf{2}$
• Yam only: $10 - (3 + 2 + 1) = 10 - 6 = \mathbf{4}$
• Plantain only: $14 - (3 + 2 + 2) = 14 - 7 = \mathbf{7}$
• Maize only: $12 - (3 + 1 + 2) = 12 - 6 = \mathbf{6}$

Summing these up: $3 + 2 + 1 + 2 + 4 + 7 + 6 = 25$.

Exam Tip

When solving Venn diagram problems, always start from the "center" (the intersection of all three sets) and work your way outwards. If the question says "only," do not subtract the triple intersection again, as it has already been excluded.