permutation & combination

[shingivyas]

Query:

I.Family consists of-an old man, 6 adults and 4 children to be seated in a row-
children wish to occupy the two seats at each end and the old man refuses to have a child on either side of him. How many arrangments possible.

Two children on eah side counted as one so in all

2 +1(old man)+6(adults) earthlings to be arranged on 9 seats.

but then old man can jus take one out of 5 seats, as he wont sit next to a child.. so that makes it 5c1- thats it am lost after that...

[Nikoloz]

first the assumptions: all are seated in one row on 11 seats, adults and children are non differentiated if switched places. Then:
4 children take the side seats and we are left just with the 6 adilts and the old chap on 7 seats. now only difference is where the old one sits. he can be seated on 5 different places (not on side seats by the children). So answer is 5.
If each child and adults are all considered different, then 4-children 4 places=4!=24. 6 adults 6 places = 6!=720 and the old one -5. so total 24*720*5=a lot!

[Nikoloz]

first the assumptions: all are seated in one row on 11 seats, adults and children are non differentiated if switched places. Then:
4 children take the side seats and we are left just with the 6 adilts and the old chap on 7 seats. now only difference is where the old one sits. he can be seated on 5 different places (not on side seats by the children). So answer is 5.
If each child and adults are all considered different, then 4-children 4 places=4!=24. 6 adults 6 places = 6!=720 and the old one -5. so total 24*720*5=a lot!

[kevinspain]

Query:

I.Family consists of-an old man, 6 adults and 4 children to be seated in a row-
children wish to occupy the two seats at each end and the old man refuses to have a child on either side of him. How many arrangments possible.

Two children on eah side counted as one so in all

2 +1(old man)+6(adults) earthlings to be arranged on 9 seats.

but then old man can jus take one out of 5 seats, as he wont sit next to a child.. so that makes it 5c1- thats it am lost after that...

Old man can sit in any of seats 3 to 9 - 7 ways to seat him
Choose two adults to sit on his left and right 6 x 5 = 30
Choose two children for seats 1 and 11 4 x 3 = 12
Other 6 people are then seated 6!

Answer 7x30x12x6!