Expert: Josh - 11/10/2004

Question
Josh,

If I have a grid 4 squares by 8 squares and 12 identical counters how many possible arrangements of these can I have assuming all the counters are on the board and cannot sit on top of one another and ignoring the fact
that some arrangements are identical if the board is rotated?

Scott.  

Answer
Hi Scott,

Under these conditions, there are C(32,12) combinations.
Where C(N,m)=N!/[(N-m)!*m!] and the factorial expression n!=n*(n-1)*(n-2)*...*3*2*1.

This class of problem is equivalent to asking how many ways we can sit "m" people down on "N" seats.

Numerically, the answer to your question is computed as
32!/[20!*12!]=(32*31*...*21)/(12*11*10*...*3*2*1)
=(1.081551316285440E17)/479001600
=225792840

There are over 225 million combinations here.
Guessing the right one is a lot more difficult than winning lotto. Considering that the odds of winning a lottory draw with 44 balls (getting six numbers right, ignoring supplementary numbers for the purpose of comparison here) is 1 in 7059052.

However, the answer may change substantially when rules are imposed.

Cheers.