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Difficulty:
95%
(hard)
Question Stats:
34% (02:39) correct
66%
(02:41)
wrong
based on 35
sessions
History
Date
Time
Result
Not Attempted Yet
Cybil, her parents, and 7 of her friends are travelling to Paris one day next week. There are two flights that day: one has 6 seats available and the other has 5. In how many different ways can these 10 people be assigned to the two flights if Cybil must be on the same flight as her parents?
A. 56
B. 70
C. 77
D. 84
E. 93
A. 56
B. 70
C. 77
D. 84
E. 93
10 Jun 2026, 14:40
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kevincan
We can take cybil and her parents as one group, say C, that has to be accommodated together..
1) Five in each flight, say X amd Y.
C in any of the flight : ways to fill up remaining two seats from seven available people = 7C2 or 21. For both X and Y, it eill be 21+21
2) Four and six respectively in each flight, say X and Y.
C in X : ways to fill up remaining one seat from seven available people = 7C1 or 7.
C in Y : ways to fill up remaining three seat from seven available people = 7C3 or 7*6*5/3! = 35.
Total ways = 21+21+7+35 = 84
11 Jun 2026, 21:37
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Dont understand where i went wrong here, would appreciate some help.
I took two situations:
1st - cybil + 2 in the flight with 6 seats
So the flight with 6 seats will have 3 friends which will be chosen as 7C3 = 35
The remaining friends will be in plane with 5 seats in 1 way.
2nd - cybil + 2 in plane with 5 seats
In same fashion, friends are chosen as 7C2 = 21
Remaining friends in the other plane, chosen in 1 way.
So answer becomes 35 + 21 = 56
Is my understanding wrong somewhere?
I took two situations:
1st - cybil + 2 in the flight with 6 seats
So the flight with 6 seats will have 3 friends which will be chosen as 7C3 = 35
The remaining friends will be in plane with 5 seats in 1 way.
2nd - cybil + 2 in plane with 5 seats
In same fashion, friends are chosen as 7C2 = 21
Remaining friends in the other plane, chosen in 1 way.
So answer becomes 35 + 21 = 56
Is my understanding wrong somewhere?
