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 A family comprised of an old man, 6 adults and 4 children is to be seated is a row with the condition that the children would occupy both the ends and never occupy either side of the old man. How many sitting arrangements are possible?

  • a)
    None 

  • b)
    4! X 5! X 6!

  • c)
    2! X 4! X 5! X 6!

  • d)
    4! X 5! X 7!

Correct answer is option 'A'. Can you explain this answer?
Verified Answer
A family comprised of an old man, 6 adults and 4 children is to be sea...
There are 11 persons and 11 seats.
Select 2 children(4C2 ways).
These two children can be seated in the side seats in 2! ways.
The old man can occupy any of the 7 middle seats(7 ways).
(note that he cannot occupy 2nd and 10th seat because these seats are near to the children sitting in the side seats)
3 seats are occupied and 8 seats are remaining. In these 8 seats, 2 seats will be near to the old man in which children cannot seat. Thus 6 seats are remaining in which 2 children can be seated. This can be done in 6P2 ways.
5 persons are seated now.  Remaining 6 persons can be arranged in the remaining 6 seats in 6! ways.
Therefore, required number of ways
= 4C2 × 2! × 7 × 6P2 ×  6!
= 6 × 2 × 7 × 30 × 720
= 1814400
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Most Upvoted Answer
A family comprised of an old man, 6 adults and 4 children is to be sea...
Given information:
- Family has an old man, 6 adults, and 4 children.
- Children occupy both ends of the row.
- Children never occupy either side of the old man.

To find: Number of possible seating arrangements.

Solution:
- Since children occupy both the ends, the old man cannot occupy the first or last position.
- Therefore, the old man can occupy any of the 8 remaining positions in the row.
- After the old man is seated, the children will occupy the first and last positions, leaving 8 positions for the rest of the family members.
- The 6 adults can be seated in any of these 8 positions in 6! ways.
- The 4 children can be seated in the remaining 2 positions in 4! ways.
- Therefore, the total number of seating arrangements = 8 x 6! x 4!

Answer: Option A, None.
Free Test
Community Answer
A family comprised of an old man, 6 adults and 4 children is to be sea...
There are 11 persons and 11 seats.

Select 2 children(4C2 ways).
These two children can be seated in the side seats in 2! ways.

The old man can occupy any of the 7 middle seats(7 ways).
(note that he cannot occupy 2nd and 10th seat because these seats are near to the children sitting in the side seats)

3 seats are occupied and 8 seats are remaining. In these 8 seats, 2 seats will be near to the old man in which children cannot seat. Thus 6 seats are remaining in which 2 children can be seated. This can be done in 6P2 ways. 

5 persons are seated now.  Remaining 6 persons can be arranged in the remaining 6 seats in 6! ways.

Therefore, required number of ways
= 4C2 x 2! x 7 x 6P2 x  6!
= 6x2x7x30x720
=1814400
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A family comprised of an old man, 6 adults and 4 children is to be seated is a row with the condition that the children would occupy both the ends and never occupy either side of the old man. How many sitting arrangements are possible?a)Noneb)4! X 5! X 6!c)2! X 4! X 5! X 6!d)4! X 5! X 7!Correct answer is option 'A'. Can you explain this answer?
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