# An olympiad level question of Combinomatrics

A family consists of a grandfather,6 sons and daughters and 4 grandchildren. They are to be seated in a row for the dinner. The grand children wish to occupy the two seats at each end and the grandfather refuses to have a grand children on either side of him. Determine the number of ways in which they can be seated for the dinner.

Note by Kumar Ashutosh
4 years, 9 months ago

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86400 is one answer received. Do anyone oppose the solution of Varnika?

- 4 years, 9 months ago

The number of ways to arrange the grand children at each end is $$4! = 4 \times 3 \times 2 \times 1 = 24$$.So the grandfather can sit in one of the 5 seats in the middle so there are 5 ways to seat the grandfather.Then the sons and daughters can be seated in $$6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 720$$.So the answer is $$24 \times 5 \times 720 = 86400$$.

- 4 years, 9 months ago

let us mark the seats as 1 2 3 4 5 6 7 8 9 10 11 children can sit on 1,2,10,11 in 4! then grandfather cannot sit on 3 and 9 so,he can arrange himself in 5 ways(5 permutation 1)and the left people can be arranged in 6!..so,total ways=4!56!=245720=86400

- 4 years, 9 months ago

great!! looks good. Thanks for your try...

- 4 years, 9 months ago