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03 Aug 2026, 09:02
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A festival organizer is putting together the final set of food trucks for a weekend street-food event. Several food trucks applied, and exactly 3 of them will be chosen. How many different groups of 3 food trucks could be chosen?
(1) If 2 more food trucks had applied, the number of possible 3-truck groups would have been 120.
(2) If 2 fewer food trucks had applied, the number of possible 3-truck groups would have been 20.
(1) If 2 more food trucks had applied, the number of possible 3-truck groups would have been 120.
(2) If 2 fewer food trucks had applied, the number of possible 3-truck groups would have been 20.
03 Aug 2026, 09:02
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Official Solution:
A festival organizer is putting together the final set of food trucks for a weekend street-food event. Several food trucks applied, and exactly 3 of them will be chosen. How many different groups of 3 food trucks could be chosen?
(1) If 2 more food trucks had applied, the number of possible 3-truck groups would have been 120.
Let \(T\) be the total number of food trucks that actually applied. If 2 more food trucks had applied, the total would have been \(T + 2\). So the number of possible 3-truck groups would have been:
\((T + 2)C3 = 120\)
\(\frac{(T + 2)!}{3!(T - 1)!} = 120\)
\(\frac{(T + 2)(T + 1)T}{6} = 120\)
There is no need to waste time trying to solve this equation algebraically. Notice that, as \(T\) increases, this expression increases. So there can be only one value of \(T\) that satisfies this equation. Therefore, statement (1) determines exactly one actual number of applicants, and hence exactly one value for the number of 3-truck groups in the original question.
Sufficient.
(2) If 2 fewer food trucks had applied, the number of possible 3-truck groups would have been 20.
If 2 fewer food trucks had applied, the total would have been \(T - 2\). So the number of possible 3-truck groups would have been:
\((T - 2)C3 = 20\)
\(\frac{(T - 2)!}{3!(T - 5)!} = 20\)
\(\frac{(T - 2)(T - 3)(T - 4)}{6} = 20\)
Again, as \(T\) increases, this expression increases. So there can be only one value of \(T\) that satisfies this equation. Therefore, statement (2) also determines exactly one actual number of applicants, and hence exactly one value for the number of 3-truck groups in the original question.
Sufficient.
Answer: D
A festival organizer is putting together the final set of food trucks for a weekend street-food event. Several food trucks applied, and exactly 3 of them will be chosen. How many different groups of 3 food trucks could be chosen?
(1) If 2 more food trucks had applied, the number of possible 3-truck groups would have been 120.
Let \(T\) be the total number of food trucks that actually applied. If 2 more food trucks had applied, the total would have been \(T + 2\). So the number of possible 3-truck groups would have been:
\((T + 2)C3 = 120\)
\(\frac{(T + 2)!}{3!(T - 1)!} = 120\)
\(\frac{(T + 2)(T + 1)T}{6} = 120\)
There is no need to waste time trying to solve this equation algebraically. Notice that, as \(T\) increases, this expression increases. So there can be only one value of \(T\) that satisfies this equation. Therefore, statement (1) determines exactly one actual number of applicants, and hence exactly one value for the number of 3-truck groups in the original question.
Sufficient.
(2) If 2 fewer food trucks had applied, the number of possible 3-truck groups would have been 20.
If 2 fewer food trucks had applied, the total would have been \(T - 2\). So the number of possible 3-truck groups would have been:
\((T - 2)C3 = 20\)
\(\frac{(T - 2)!}{3!(T - 5)!} = 20\)
\(\frac{(T - 2)(T - 3)(T - 4)}{6} = 20\)
Again, as \(T\) increases, this expression increases. So there can be only one value of \(T\) that satisfies this equation. Therefore, statement (2) also determines exactly one actual number of applicants, and hence exactly one value for the number of 3-truck groups in the original question.
Sufficient.
Answer: D
