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17 Jul 2026, 08:00
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C
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Difficulty:
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68% (02:07) correct
32%
(02:18)
wrong
based on 215
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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.
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M46-58
17 Jul 2026, 09:00
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GMAT Club 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 (1), we know:
Drama = 5
Chess = 4
Robotics = 3
Art = 2
These account for 14 volunteers, so 3 volunteers remain.
From (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
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 (1), we know:
Drama = 5
Chess = 4
Robotics = 3
Art = 2
These account for 14 volunteers, so 3 volunteers remain.
From (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
General Discussion
17 Jul 2026, 08:28
Kudos
Bookmarks
Given
17 V, every V is in a club
At least 3 clubs were there
DC had 5 V, CC had 4 V
other clubs should have 8 V (the left over)
we need how many ways one can select 3 volunteers, from different clubs, none to be from the same club
so we need to choose 0 or atmost 1 from x number of clubs
S1 - RC had 3 V and AC had 2 V
meaning of the 8, 5 were in the two clubs mentioned above, the rest 3 can be in three clubs, or two clubs or one club
so the ways will always vary with the number of clubs
Insufficient
S2 - Exactly one club has less than 3, so the club could either have 1 or 2 V, and the remaining 7 or 6 V are still needed to be divided to unknown number of clubs, so also Insufficient
S1+S2 = Exactly one club has less than 3, that is AC mentioned in S1, which has 2 V, so of the left over 3 volunteers, the only possible way is for all three of them to be in one club say X club
so answer could be derived, as there are 5 clubs a fixed number
(sufficient)
Ans - C
17 V, every V is in a club
At least 3 clubs were there
DC had 5 V, CC had 4 V
other clubs should have 8 V (the left over)
we need how many ways one can select 3 volunteers, from different clubs, none to be from the same club
so we need to choose 0 or atmost 1 from x number of clubs
S1 - RC had 3 V and AC had 2 V
meaning of the 8, 5 were in the two clubs mentioned above, the rest 3 can be in three clubs, or two clubs or one club
so the ways will always vary with the number of clubs
Insufficient
S2 - Exactly one club has less than 3, so the club could either have 1 or 2 V, and the remaining 7 or 6 V are still needed to be divided to unknown number of clubs, so also Insufficient
S1+S2 = Exactly one club has less than 3, that is AC mentioned in S1, which has 2 V, so of the left over 3 volunteers, the only possible way is for all three of them to be in one club say X club
so answer could be derived, as there are 5 clubs a fixed number
(sufficient)
Ans - C
17 Jul 2026, 09:02
Kudos
Bookmarks
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.
We don't know remaining 3 volunteers were from same club or different.
Insufficient
(2) Among the 17 volunteers, exactly one club was represented by fewer than 3 volunteers.
Remaining 8 volunteers distribution is not clear, many possibilities.
Insufficient
(1)&(2)
Total =17
Drama club=5
Chess Club=4
Robotics Club= 3
Art Club= 2
The 3 remaining volunteers must be a part of same club other than the ones we are given information about.
17 volunteers are in clubs of sizes 5,4,3,3,2.
We have unique values of all club sizes, now we can easily find the ways in which team can be formed.
Total invalid ways= 5C2(12)+5C3+4C2(13)+4C3+3C2(14)+3C3+2C2(15)+3C2(14)+3C3= 120+10+78+4+43+43+15= 313
Valid number of ways= 680-313= 367
Sufficient
C
(1) Among the 17 volunteers, 3 represented the Robotics Club and 2 represented the Art Club.
We don't know remaining 3 volunteers were from same club or different.
Insufficient
(2) Among the 17 volunteers, exactly one club was represented by fewer than 3 volunteers.
Remaining 8 volunteers distribution is not clear, many possibilities.
Insufficient
(1)&(2)
Total =17
Drama club=5
Chess Club=4
Robotics Club= 3
Art Club= 2
The 3 remaining volunteers must be a part of same club other than the ones we are given information about.
17 volunteers are in clubs of sizes 5,4,3,3,2.
We have unique values of all club sizes, now we can easily find the ways in which team can be formed.
Total invalid ways= 5C2(12)+5C3+4C2(13)+4C3+3C2(14)+3C3+2C2(15)+3C2(14)+3C3= 120+10+78+4+43+43+15= 313
Valid number of ways= 680-313= 367
Sufficient
C
