In a group of players in a college, 20 are in the basketball team, 24 in the hockey team and 27 in the cricket team. If 12 play hockey and basketball, 10 play cricket and basketball, 14 play hockey and cricket and 7 play all the three games, then the total number of players in the group is
Aptitude
Simplification
Difficulty: Medium
Choose an option
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A42
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B43
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C45
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D49
Answer
Correct Answer: 42
Explanation
### Concept & Logic
This problem requires the three-set version of the Inclusion-Exclusion Principle.
When you add the totals of three overlapping circles, you double-count the pairwise intersections, and triple-count the center intersection. The formula corrects for this by adding the singles, subtracting the doubles, and re-adding the triple overlap.
### Step-by-Step Solution
* **Given:** Basketball, $n(B) = 20$
Hockey, $n(H) = 24$
Cricket, $n(C) = 27$
Hockey & Basketball, $n(H \cap B) = 12$
Cricket & Basketball, $n(C \cap B) = 10$
Hockey & Cricket, $n(H \cap C) = 14$
All three, $n(B \cap H \cap C) = 7$
* **Calculation:** * Use the grand formula for the union of three sets:
$$n(B \cup H \cup C) = n(B) + n(H) + n(C) - n(H \cap B) - n(C \cap B) - n(H \cap C) + n(B \cap H \cap C)$$
* Substitute the provided numbers into the equation:
$$\text{Total Players} = 20 + 24 + 27 - 12 - 10 - 14 + 7$$
* Group the positive and negative numbers for easier mental math:
$$\text{Positives} = 20 + 24 + 27 + 7 = 78$$
$$\text{Negatives} = 12 + 10 + 14 = 36$$
* Subtract the totals:
$$\text{Total Players} = 78 - 36 = 42$$
### Exam Strategy & Shortcut
You can calculate this rapidly by treating the positive and negative pairs together to keep numbers small.
$(20 - 12 = 8)$
$(24 - 10 = 14)$
$(27 - 14 = 13)$
Add these up with the center overlap: $8 + 14 + 13 + 7 = 42$.
This minimizes the risk of arithmetic errors associated with summing large strings of numbers.
### Common Pitfall
The most dangerous pitfall is forgetting to *add back* the 7 players who play all three games at the end of the formula. Because they were subtracted three times (in each of the pairwise intersections), they have been entirely removed from the total count and must be reinstated.
### Final Answer
**Therefore, the correct answer is 42.**