First let's choose any 3 of the 9 friends to place at the table
which would be C(9,3) or 84
Now it depends on what we consider the seating arrangement at the
table.
If the seats are specified, e.g. one facing against a wall , etc
the each of the groups of 84 triples can be arranged in 3! or 6 ways.
So the number of seatings is 84*6 or 504
However, suppose everybody stands up and moves one seat to the left.
Is the seating arrangement still the same? e.g. for a round table it would still
be the same.
The given answer seems to suggest that. So let's place one of the chosen
people at the table, then the other 2 can be seated in 2*1 or 2 ways.
So.... number of ways = 84*3! /2 = 252
I agree, it is a rather tricky one.
A group of 9 friends go into a nightclub and there is only a table for 3 and therefore the others will have to stand up. If the friends randomly assign who sits at the table, in how many ways can they be seated?
These permutation and combination q's are giving me such a hard time Thanks so much to anyone who helps. Answer is 252
1 answer