Seated Row
March 7th, 2010
Seated Row
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A family comprised of an old man , 6 adults and 4 children is to be seated in a row with the condition that?
the children would occupy both the ends and never occupy either side of the the old man.how many seating arrangements are possible.
look at the scheme below [ C = child, O = old man]
C x O O O O O O O x C
2 children have to occupy the ends,
leaving 7 possible seats for the old man
who can't occupy the seats marked x
two cases can now arise
case (a): old man seated towards the end [2 seats]
C x O x - - - - - - C
x now indicates where children can't be seated
so the remaining 2 children have 6 possible places to sit
case (b): old man not seated at the ends [5 seats]
C x x O x - - - - - C
so the remaining 2 children have 5 possible places to sit
we can now work out all the permutations
(1)
4P2 = 12 [2 children chosen for end seats & permutated]
2(a)
2 = [old man allocated a seat]
6P2 = 30 [remaining 2 children seated & permutated]
2(b)
5 = [old man allocated a seat]
5P2 = 20 [remaining 2 children seated & permutated]
(3)
6P6 = 720 [6 adults permutated in 6 seats left]
(4)
thus # of seating arrangements
= 4P2*[2*30 + 5*20]*6P6
= 12*160*720
= 1,382,400