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09 Sep 2024, 01:30
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
85%
(hard)
Question Stats:
39% (02:31) correct
61%
(02:42)
wrong
based on 76
sessions
History
Date
Time
Result
Not Attempted Yet
A treasure hunt competition requires a team of 4 students. There are 6 students who are willing to participate: John, Edward, Milner, Luiz, Howard and Paul. However, Edward and Milner do not want to be on a team together, while Paul only wants to be on a team if Edward is also. How many different teams can be formed?
A. 3
B. 5
C. 6
D. 7
E. 8
A. 3
B. 5
C. 6
D. 7
E. 8
09 Sep 2024, 02:25
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If E is on the team, then J, L, H,P anyone can be in the team.
Total combinations that can be formed with E,J,L,H,P = 4 (because if E is not in the team, then P is also not in the team, however the reverse is not true.) (EPJL, EPJH, EPLH, EJLH)
If M is on the team, then E is not in the team and hence P is also not in the team
So, with M on the team only 1 team is possible. (M,J,L,H)
Hence, total teams that can be formed according to given constraints = 4+1=5
B is CORRECT.
Posted from my mobile device
Total combinations that can be formed with E,J,L,H,P = 4 (because if E is not in the team, then P is also not in the team, however the reverse is not true.) (EPJL, EPJH, EPLH, EJLH)
If M is on the team, then E is not in the team and hence P is also not in the team
So, with M on the team only 1 team is possible. (M,J,L,H)
Hence, total teams that can be formed according to given constraints = 4+1=5
B is CORRECT.
Posted from my mobile device
09 Sep 2024, 05:46
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Case 1: E on team and M not
As we need to select 3 from the remaining 4 people; 4C3 = 4
Case 2: M on team and E not => also P not
As we need to select 3 from the remaining 3 people; 3C3 = 1
No other case is possible, both can't be together and if both of them are not there then even P won't be there and team of 4 can't be formed.
Therefore, 4 + 1 = 5
Answer B.
As we need to select 3 from the remaining 4 people; 4C3 = 4
Case 2: M on team and E not => also P not
As we need to select 3 from the remaining 3 people; 3C3 = 1
No other case is possible, both can't be together and if both of them are not there then even P won't be there and team of 4 can't be formed.
Therefore, 4 + 1 = 5
Answer B.
