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07 Aug 2026, 10:30
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At a school festival, 17 students volunteered to lead activity booths. Each volunteer represented exactly one school club, and the volunteers represented at least 3 clubs. Among the volunteers, 5 represented the Drama Club and 4 represented the Chess Club. If a 3-student team is to be formed so that no two team members represent the same club, in how many ways can the team be formed?
(1) Among the 17 volunteers, 3 represented the Robotics Club and 2 represented the Art Club.
(2) Among the 17 volunteers, exactly one club was represented by fewer than 3 volunteers.
(1) Among the 17 volunteers, 3 represented the Robotics Club and 2 represented the Art Club.
(2) Among the 17 volunteers, exactly one club was represented by fewer than 3 volunteers.
07 Aug 2026, 10:30
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Official Solution:
At a school festival, 17 students volunteered to lead activity booths. Each volunteer represented exactly one school club, and the volunteers represented at least 3 clubs. Among the volunteers, 5 represented the Drama Club and 4 represented the Chess Club. If a 3-student team is to be formed so that no two team members represent the same club, in how many ways can the team be formed?
(1) Among the 17 volunteers, 3 represented the Robotics Club and 2 represented the Art Club.
Now we know:
• Drama \(= 5\)
• Chess \(= 4\)
• Robotics \(= 3\)
• Art \(= 2\)
These account for 14 volunteers, so 3 volunteers remain.
But those 3 remaining volunteers could all represent one additional club, giving club counts 5, 4, 3, 2, and 3.
Or they could be split between two additional clubs, giving club counts 5, 4, 3, 2, 2, and 1.
These distributions give different numbers of possible teams.
Not sufficient.
(2) Among the 17 volunteers, exactly one club was represented by fewer than 3 volunteers.
This restricts the distribution, but it does not determine the distribution.
For example, after the 5 Drama Club volunteers and the 4 Chess Club volunteers, the remaining 8 volunteers could be split among other clubs as 2 and 6, giving club counts 5, 4, 2, and 6.
Or the remaining 8 volunteers could be split as 2, 3, and 3, giving club counts 5, 4, 2, 3, and 3.
Both cases satisfy statement (2), but they give different numbers of possible teams.
Not sufficient.
(1)+(2) From statement (1), we know:
• Drama \(= 5\)
• Chess \(= 4\)
• Robotics \(= 3\)
• Art \(= 2\)
These account for 14 volunteers, so 3 volunteers remain.
From statement (2), exactly one club was represented by fewer than 3 volunteers. Since the Art Club already has 2 volunteers, the remaining 3 volunteers cannot be split among two or more additional clubs, because that would create another club with fewer than 3 volunteers.
Therefore, the remaining 3 volunteers must all represent one additional club. So the club counts are:
\(5, 4, 3, 2, 3\)
So the distribution is fixed, and we can determine the number of possible 3-student teams.
Sufficient.
Answer: C.
For reference, to count the possible 3-student teams, first choose 3 different clubs to delegate one student each to the team, and then choose 1 student from each selected club.
The club counts are:
• Drama \(= 5\)
• Chess \(= 4\)
• Robotics \(= 3\)
• Art \(= 2\)
• Other club \(= 3\)
Now count each possible choice of 3 clubs:
• Drama, Chess, Robotics: \(5 * 4 * 3 = 60\)
• Drama, Chess, Art: \(5 * 4 * 2 = 40\)
• Drama, Chess, Other club: \(5 * 4 * 3 = 60\)
• Drama, Robotics, Art: \(5 * 3 * 2 = 30\)
• Drama, Robotics, Other club: \(5 * 3 * 3 = 45\)
• Drama, Art, Other club: \(5 * 2 * 3 = 30\)
• Chess, Robotics, Art: \(4 * 3 * 2 = 24\)
• Chess, Robotics, Other club: \(4 * 3 * 3 = 36\)
• Chess, Art, Other club: \(4 * 2 * 3 = 24\)
• Robotics, Art, Other club: \(3 * 2 * 3 = 18\)
Total:
\(60 + 40 + 60 + 30 + 45 + 30 + 24 + 36 + 24 + 18 = 367\)
Answer: C
At a school festival, 17 students volunteered to lead activity booths. Each volunteer represented exactly one school club, and the volunteers represented at least 3 clubs. Among the volunteers, 5 represented the Drama Club and 4 represented the Chess Club. If a 3-student team is to be formed so that no two team members represent the same club, in how many ways can the team be formed?
(1) Among the 17 volunteers, 3 represented the Robotics Club and 2 represented the Art Club.
Now we know:
• Drama \(= 5\)
• Chess \(= 4\)
• Robotics \(= 3\)
• Art \(= 2\)
These account for 14 volunteers, so 3 volunteers remain.
But those 3 remaining volunteers could all represent one additional club, giving club counts 5, 4, 3, 2, and 3.
Or they could be split between two additional clubs, giving club counts 5, 4, 3, 2, 2, and 1.
These distributions give different numbers of possible teams.
Not sufficient.
(2) Among the 17 volunteers, exactly one club was represented by fewer than 3 volunteers.
This restricts the distribution, but it does not determine the distribution.
For example, after the 5 Drama Club volunteers and the 4 Chess Club volunteers, the remaining 8 volunteers could be split among other clubs as 2 and 6, giving club counts 5, 4, 2, and 6.
Or the remaining 8 volunteers could be split as 2, 3, and 3, giving club counts 5, 4, 2, 3, and 3.
Both cases satisfy statement (2), but they give different numbers of possible teams.
Not sufficient.
(1)+(2) From statement (1), we know:
• Drama \(= 5\)
• Chess \(= 4\)
• Robotics \(= 3\)
• Art \(= 2\)
These account for 14 volunteers, so 3 volunteers remain.
From statement (2), exactly one club was represented by fewer than 3 volunteers. Since the Art Club already has 2 volunteers, the remaining 3 volunteers cannot be split among two or more additional clubs, because that would create another club with fewer than 3 volunteers.
Therefore, the remaining 3 volunteers must all represent one additional club. So the club counts are:
\(5, 4, 3, 2, 3\)
So the distribution is fixed, and we can determine the number of possible 3-student teams.
Sufficient.
Answer: C.
For reference, to count the possible 3-student teams, first choose 3 different clubs to delegate one student each to the team, and then choose 1 student from each selected club.
The club counts are:
• Drama \(= 5\)
• Chess \(= 4\)
• Robotics \(= 3\)
• Art \(= 2\)
• Other club \(= 3\)
Now count each possible choice of 3 clubs:
• Drama, Chess, Robotics: \(5 * 4 * 3 = 60\)
• Drama, Chess, Art: \(5 * 4 * 2 = 40\)
• Drama, Chess, Other club: \(5 * 4 * 3 = 60\)
• Drama, Robotics, Art: \(5 * 3 * 2 = 30\)
• Drama, Robotics, Other club: \(5 * 3 * 3 = 45\)
• Drama, Art, Other club: \(5 * 2 * 3 = 30\)
• Chess, Robotics, Art: \(4 * 3 * 2 = 24\)
• Chess, Robotics, Other club: \(4 * 3 * 3 = 36\)
• Chess, Art, Other club: \(4 * 2 * 3 = 24\)
• Robotics, Art, Other club: \(3 * 2 * 3 = 18\)
Total:
\(60 + 40 + 60 + 30 + 45 + 30 + 24 + 36 + 24 + 18 = 367\)
Answer: C
