Please HELP me ASAP

Question

"Sixteen people are sitting around a circular table, each holding a fair coin. All sixteen people flip their coins and those who flip heads stand while those who flip tals remain seated. What is the probability that no two adjacent people will stand?

Answer 1

As there are only 65536, and only 16 loops of 0 to 1 necessary, you can calculate this by brute force, checking for each of the 16 if they are sitting or both of the people next to them or sitting, and whether this is true for everyone.

I'm sure that there is a more elegant way to solve this. However, one can use combinatoric arguments to give the number of ways to have no adjacent people standing with 0 to 8 people standing.

There is one way to have no one standing, 16 ways to have one person standing, and, for 2 through 8, since there must be a gap of at least one between each person standing, you can subtract s-1 from the s standing. Thus, we have C(16-(s-1),s) = C(16-s+1,s) = C(17-s, s). Also, though, you must remove the count with someone standing in the first and last position and not standing in the second and next to last position equals C(12-((s-2)-1),s-2) = C(15-s, s-2) . Thus, it is C(17 - s, s) - C(15 - s, s-2) for s = 2 to 8

Thus, we have, writing the number standing and the number of ways as an ordered pair

(0, 1), (1, 16), (2, 104), (3, 352), (4, 660), (5, 672), (6, 336), (7, 64), (8, 2)

This sums to 2207 ways.

Then, the probability is 2207/2^16 = 2207/65536.