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18 Jul 2026, 03:44
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Bookmarks
From the Stem we know clubs are at least 3 and that Drama has 5 and Chess 4
S1 We get that 3 more are from robotics and 2 from art = meaning total is 5+4+3+2= 14 meaning we still have 3 unaccounted for hence insufficient
S2 We already know Drama is 5 and Chess 4 so there must be other clubs with less than 3 but we do not know the distribution still so insufficient
S1+S2 Combined We have
Drama 5
Chess 4
Robotics 3
Art 2
Other club 3 (has to be three since only one club has less than 3)
Now we can safely say we have 5 clubs with the distribution shown above which is enough to find the team of 3
Ans C
S1 We get that 3 more are from robotics and 2 from art = meaning total is 5+4+3+2= 14 meaning we still have 3 unaccounted for hence insufficient
S2 We already know Drama is 5 and Chess 4 so there must be other clubs with less than 3 but we do not know the distribution still so insufficient
S1+S2 Combined We have
Drama 5
Chess 4
Robotics 3
Art 2
Other club 3 (has to be three since only one club has less than 3)
Now we can safely say we have 5 clubs with the distribution shown above which is enough to find the team of 3
Ans C
Bunuel
18 Jul 2026, 03:50
Kudos
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Both statements together are sufficient but neither alone is sufficient
Sol
17 students , at least 3 clubs
5 - drama
4- chess
To form 3-student team with no two team members on the same club
Statement 1 -
3 - Robotics
2- Art club
So Total - 5+4+3+2 =14
So the rest of the students can be either in
1 club - 3 students
3 different clubs - 1 student each
2 clubs - 2 in one and 1 in another
This does not provide any valid sol so this is not sufficient
Statement 2
Exactly 1 club has <3 students
5 in d and 4 in chess
remaining 8 students can be in n no of possible options so this is cannot be determined
Using Statement 1 and 2
Total - 14
Remaining - 3
Exactly 1 <3 so we already have it as Art club that has only 2 students
So we know other groups can only have >3 so , 1 more group can only be made
So total clubs = 5
From this we can fins the possible ways
Teams = (5
3) = 10
Finding the 10 combinations
5x4x3 +5x4X3 + 5x4x2 + 5x3x3 + 3x3x2 + 4x3x2 + 4x3x2 + 4x3x3 + 5x3x2 +5x3x2 = 367 ways
Sol
17 students , at least 3 clubs
5 - drama
4- chess
To form 3-student team with no two team members on the same club
Statement 1 -
3 - Robotics
2- Art club
So Total - 5+4+3+2 =14
So the rest of the students can be either in
1 club - 3 students
3 different clubs - 1 student each
2 clubs - 2 in one and 1 in another
This does not provide any valid sol so this is not sufficient
Statement 2
Exactly 1 club has <3 students
5 in d and 4 in chess
remaining 8 students can be in n no of possible options so this is cannot be determined
Using Statement 1 and 2
Total - 14
Remaining - 3
Exactly 1 <3 so we already have it as Art club that has only 2 students
So we know other groups can only have >3 so , 1 more group can only be made
So total clubs = 5
From this we can fins the possible ways
Teams = (5
3) = 10
Finding the 10 combinations
5x4x3 +5x4X3 + 5x4x2 + 5x3x3 + 3x3x2 + 4x3x2 + 4x3x2 + 4x3x3 + 5x3x2 +5x3x2 = 367 ways
Bunuel
18 Jul 2026, 04:16
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Bookmarks
Ans Choice C: Given Total volunteers = 17 ; No overlap of volunteers btw clubs. ; Drama = 5 ; Chess = 4
In order to cal ways to form a 3 student team so that no two team menbers represent same club, we need to know exactly how many clubs are there apart from Drama, Chess.
Stmt1: Robotics = 3 ; Art = 2
But this still leaves for 3 unaccounted volunteers; which can be part of 1,2,3 member clubs
Not Suff.
Stmt2: Exactly 1 Club < 3 (i.e 1, 2 volunteers)
Case 1 with exactly 1 Club with 1 volunteer will lead to cases with single club with 7 members, 2clubs with 3 & 4 members.
Case 2 with exactly 1 club with 2 vounteers will lead to cases with single club with 6 members, 2 clubs with 3 & 3 members.
Not Suff.
Combining 1 + 2 : We get Drama=5; Chess=4; Robotics=3 ; Art=2 and exactly 1 club with < 3 members
Since Art club=2 <3
The other club with 3 members should be part of a single club, Club x = 3
Now the no. of ways of selecting a 3 member team from a total of 5 clubs can be found out.
Sufficient.
In order to cal ways to form a 3 student team so that no two team menbers represent same club, we need to know exactly how many clubs are there apart from Drama, Chess.
Stmt1: Robotics = 3 ; Art = 2
But this still leaves for 3 unaccounted volunteers; which can be part of 1,2,3 member clubs
Not Suff.
Stmt2: Exactly 1 Club < 3 (i.e 1, 2 volunteers)
Case 1 with exactly 1 Club with 1 volunteer will lead to cases with single club with 7 members, 2clubs with 3 & 4 members.
Case 2 with exactly 1 club with 2 vounteers will lead to cases with single club with 6 members, 2 clubs with 3 & 3 members.
Not Suff.
Combining 1 + 2 : We get Drama=5; Chess=4; Robotics=3 ; Art=2 and exactly 1 club with < 3 members
Since Art club=2 <3
The other club with 3 members should be part of a single club, Club x = 3
Now the no. of ways of selecting a 3 member team from a total of 5 clubs can be found out.
Sufficient.
Bunuel
