Enael wrote:
Does the formula 6!/3!3! only work if they are sitting next to each other?
e.g. sitting DDDEEE or EEEDDD.
Or is it due to having 2 different sub-sets that we can't use such formula.
Thank you for the help.
6!/(3!3!)=20 is the number of arrangements of 6 letters EEEDDD, where 3 D's and 3 E's areidentical :
EEEDDD;
EEDEDD;
EDEEDD;
DEEEDD;
...
DDDEEE.
But in our original question we don't have 3 identical D's and 3 identical E's. Also, we need only those arrangements where no dwarf will sit
...