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.

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Easy Math Editor

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`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
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Comments

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

Kumar Ashutosh - 7 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

varnika chaturvedi - 7 years, 9 months ago

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

Kumar Ashutosh - 7 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 $24 \times 5 \times 720 = 86400$ .

Tan Li Xuan - 7 years, 9 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$