Math, asked by nidhi918, 7 months ago

95. 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?
need explanation​

Answers

Answered by AditiHegde
15

Given:

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.  

To find:

How many sitting arrangements are possible?

Solution:

A family comprised of an old man, 6 adults and 4 children.

1 + 6 + 4 = 11 persons and similarly 11 seats.

Children would occupy both the ends and never occupy either side of the old man.  

There are 4C2 ways to select 2 children.

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).

An old man cannot occupy 2nd and 10th seat as they are by the side of seats of children.

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 is given by 6P2 ways.

5 persons are seated now.  

Remaining 6 persons can be arranged in the remaining 6 seats in 6! ways.

Therefore, the required number of ways

= 4C2 × 2! × 7 × 6P2 ×  6!

= 6 × 2 × 7 × 30 × 720

= 1814400

Therefore, 1814400 sitting arrangements are possible.

Answered by Aditiiiiiiiiiii
11

(a) 4!×5!×7!

CHILDREN (4)

First we will arrange children as they have to occupy only end seats (1st,2nd,10th&11th) & they are 4 in number so 4!

OLD MAN (1)

then secondly we will arrange old Man as 4 seats (1st,2nd,10th& 11th) are already occupied & he cannot sit on either side of children which means he cannot sit on seats number 3rd & 9th so now he only have 5 seats left to choose (4th,5th,6th,7th&8th) i.e. 5!

ADULTS (7)

lastly we will see Adults As seats (1st,2nd,10th&11th) are already occupied by children. They have 7 seats left (3rd,4th,5th,6th,7th,8th&9th) to choose so 7!

Important Part of this question is that In Adult case old Man will not be considered fixed (like children) because adults & old man may permute between seats (4th,5th,6th&7th)

I hope you will Understand my Answer

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