No, suppose you take 4 different types of flowers namely a,b,c,d for the first 4 tables respectively. (i.e. a on first, b on second, c on third, d on fourth).
Then the combinations for the latter 4 tables will be like:
For table 5: Suppose a combination of a,b,c
For table 6: a combination of b,c,d
For table 7: a combination of c,d,a
For table 8: a combination of a,c,d
Note that at least one flower type is different in each combination which makes it unique.
AKG1593
I don't understand the explanation.How do we arrive at the equation:\(n+C(n,3)=8\)?
If we take 8 different types of single flower bouquet then only we can have each table unique.
On the other hand,if we take 4 different types of flowers and decorate 4 tables with single flowers and the other four with different combinations of the 4 types of flowers won't the latter 4 tables essentially be familiar as the type of flowers remain the same irrespective of how we arrange them?
Bunuel,could you take on this one please?