Question

an experiment consists of tossing a coin eleven times and the sequence of heads and tails is observed. how many of the possible outcomes contain five heads, with no two heads adjacent to each other?

Answer 1

To count the number of outcomes where there are five heads and no two of them are adjacent, we can use the following approach:

Let H denote a head and T denote a tail. We need to arrange 5 H's and 6 T's in such a way that no two H's are adjacent. We can start by placing the 6 T's in a row as shown below:

T _ T _ T _ T _ T _ T

The 5 H's must be placed in the 6 gaps indicated by the blanks. However, we cannot place two H's in the same gap, and we cannot place an H at either end of the row, since that would make it adjacent to a T. Therefore, we need to choose 5 gaps out of the 6 gaps available, and place one H in each of the chosen gaps. This can be done in:

C(6,5) x P(5,5) = 6 x 5! = 720

ways, where C(6,5) is the number of ways to choose 5 gaps out of 6, and P(5,5) is the number of ways to arrange the 5 H's in the chosen gaps.

Finally, we can observe that each toss of the coin can result in 2 possible outcomes (H or T), so the total number of outcomes of 11 coin tosses is 2^11 = 2048.

Therefore, the number of possible outcomes where there are exactly 5 heads and no two of them are adjacent is:

720 x 2^6 = 720 x 64 = 46,080

So there are 46,080 possible outcomes of 11 coin tosses where there are exactly 5 heads and no two of them are adjacent.