In a restaurant there are four identical circular tables. each tables have 4 chairs. Two families of 7 members and 8 members come to the restaurant. Then find the number of ways these members can be seated (if members of two families do not sit on the same table).





My approach selection of members into four groups $^8C_4*^7C_4$ .This forms numbers of ways of four distinct groups. Now four group are selected . Lets treat the four group as 4 distinct balls and number of tables as 4 identical balls as the word identical is clearly mentioned but I am not able to solve it


1 Answer 1


there's a catch when you divide 8 people in 2 groups of four then you have to divide ${8 \choose 4}$ by 2 (let's say you choose 1,2,3,4 but that's identical with choosing 5,6,7,8).

Now if each table is identical , it does not matter where family members go to, we will say the ways to allocate the tables to 4 groups is '1'.

The ways of arranging 3 groups of 4 people is $(3!)^3$(ways to arrange 4 people on a circle is 3!).

As for the other group with 3 people can be treated 3 people and 1 hole (it matters which 2 persons are around hole),thus making it the case of arranging 4 objects on a circle which again is 3!.

our answer would be $\frac{7!8!3!3!3!3!}{4!4!4!3!2!}=\frac{7!7!\times8}{2^7}=\frac{(7!)^2}{16}$.

sorry if it's a little not understandable.


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