Solve

The answer is 1814400

first two children should be chosen from 4 children to occupy the end seats. this can be done in 6 ways. Now they can be seated in 2 ways.

next we select a seat for the old man. he can't sit in 2nd seat or 10th seat as they are beside the end seats in which children sat. He can sit anywhere from 3rd seat to 9th seat. there are 7 seats so there are 7 ways in which he can be seated.

now in the remaining 8 seats the remaining children can't occupy two seats that are beside the old man. therefore they should sit in the remaining 6 seats which can be done in 6P2 ways i.e 30 ways.

now there are 6 seats remaining. 6 adults can be arranged in those 6 seats in 6! ways i.e. 720 ways.

hence the answer is 6*2*7*30*720=18144000

ans. 15.

1c1*6c2*4c4 = 15

I can think of another way to solve the problem.(similar to the way that is given in illustrations of the book)

The old man has to sit with two adults on the either side of him. now we will create a group with old man and two adults and make him a single entity. This can be done in 6C2*2 ways. (selecting 2 adults from the 6 adults and placing them in 2 sides of the old man) i.e. 30 ways. now we can consider  such a group as one adult as the group can be place anywhere except at the end. let us chose 2 children to be seated in ends this can be done in 6 ways and they can be seated in 2 ways. now the remaing 5 adults( 4 adults remaining from the selection of the group and one group treated as an adult) and  2 children(total 7) can be arranged without any restrictions in 7! ways. the answer is 7!x(4!/2!)x(6!/4!) We get the same answer.