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01 Sep 2022, 04:40
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
75%
(hard)
Question Stats:
67% (02:58) correct
33%
(02:57)
wrong
based on 170
sessions
History
Date
Time
Result
Not Attempted Yet
A family including a father, a mother, three sons and three daughters, decide to watch a movie in their home theater room. The theater room has two front seats, three seats in the second row and three seats in the third row. The right most seat in the front row is reserved for one of the parents. If all the daughters want to sit in the same row and two of their sons, William and Jacob refuse to sit next to each other, in how many ways can the family be seated in the home theater room?
A. 320
B. 380
C. 384
D. 390
E. 400
A. 320
B. 380
C. 384
D. 390
E. 400
01 Sep 2022, 23:11
Kudos
Bookmarks
Since the rightmost seat of the first row is already reserved for a parent, it can be filled in two ways.
Also, as the girls have to sit together, they will occupy one whole row and hence can be seated in 2*3! ways. 3! is multiplied for the arrangement of the girls among themselves.
Now, they are left with 4 vacant seats. (1 in the first row and 3 in one of the vacant rows)
Let us consider the cases for William.
1) If he is sitting in the first row, the number of ways the other three will sit will be 3! = 6.
2)William is sitting in one of the vacant rows:
I)W_ _: For this arrangement, the middle seat of the vacant row can be filled in two ways, and the other 2 seats can also be filled in two ways.
Thus, number of arrangements in this case = 2*2 = 4.
Similarly, for _ _ W, the number of arrangements will be 4.
II)_ W _: Jacob will sit in the front row for this arrangement. The total number of arrangements will be 2.
Thus, the total number of arrangements = 2*2*3!(3! + 4 + 4 + 2) = 24 * 16 = 384
Thus, the correct option is C.
Also, as the girls have to sit together, they will occupy one whole row and hence can be seated in 2*3! ways. 3! is multiplied for the arrangement of the girls among themselves.
Now, they are left with 4 vacant seats. (1 in the first row and 3 in one of the vacant rows)
Let us consider the cases for William.
1) If he is sitting in the first row, the number of ways the other three will sit will be 3! = 6.
2)William is sitting in one of the vacant rows:
I)W_ _: For this arrangement, the middle seat of the vacant row can be filled in two ways, and the other 2 seats can also be filled in two ways.
Thus, number of arrangements in this case = 2*2 = 4.
Similarly, for _ _ W, the number of arrangements will be 4.
II)_ W _: Jacob will sit in the front row for this arrangement. The total number of arrangements will be 2.
Thus, the total number of arrangements = 2*2*3!(3! + 4 + 4 + 2) = 24 * 16 = 384
Thus, the correct option is C.
01 Sep 2022, 07:37
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IMO the correct answer is "C" 384
My calculations are as follows
2 * 2 * 6 * (2 * 6 + 2 * 2)
2 ways of choosing the parent seated in the rightmost front seat (mother or father)
2 ways of choosing the row where the three daughters are seated (as a matter of fact, it could be the second or the third row)
6 ways of arranging the sisters (3!)
...times...
the sum of two different cases:
1) Either W or J are seated in the leftmost seat in the front row
Then we would have
2 ways of choosing one of them (W or J)
6 ways of arranging the remaining members of the family (that is, the one between W or J and the one between the two parents who have not previously been chosen as being seated in the front row, plus the other son) in the three-seat row we are left with
2) Neither W or J are seated in the leftmost seat in the front row and they are therefore sharing the same not-occupied-by-their-sisters three-seat row
As a consequence of the constraint which impose the two of them NOT to be seated next to each other, another family member must be seated in between
Then we would have
2 ways to choose this family member (who could essentially be the other son or the other parent who is not seated in the front row)
2 ways of arranging W and J (as a matter of fact, that row could be in the form "W-other family member-J" or "J-other family member-W")
In this way, I think all possible configurations have been exhausted and taken into account
In the end, the calculations boil down to
24 * 16 = 384
Answer "C"
Posted from my mobile device
My calculations are as follows
2 * 2 * 6 * (2 * 6 + 2 * 2)
2 ways of choosing the parent seated in the rightmost front seat (mother or father)
2 ways of choosing the row where the three daughters are seated (as a matter of fact, it could be the second or the third row)
6 ways of arranging the sisters (3!)
...times...
the sum of two different cases:
1) Either W or J are seated in the leftmost seat in the front row
Then we would have
2 ways of choosing one of them (W or J)
6 ways of arranging the remaining members of the family (that is, the one between W or J and the one between the two parents who have not previously been chosen as being seated in the front row, plus the other son) in the three-seat row we are left with
2) Neither W or J are seated in the leftmost seat in the front row and they are therefore sharing the same not-occupied-by-their-sisters three-seat row
As a consequence of the constraint which impose the two of them NOT to be seated next to each other, another family member must be seated in between
Then we would have
2 ways to choose this family member (who could essentially be the other son or the other parent who is not seated in the front row)
2 ways of arranging W and J (as a matter of fact, that row could be in the form "W-other family member-J" or "J-other family member-W")
In this way, I think all possible configurations have been exhausted and taken into account
In the end, the calculations boil down to
24 * 16 = 384
Answer "C"
Posted from my mobile device
