To solve this problem, we need to determine how many possible outcomes contain exactly five heads (H) and no two heads are adjacent, out of a total of fifteen tosses of a coin.

Step 1: Understanding the problem

• There are 15 coin tosses.
• We need to place exactly 5 heads (H) such that no two heads are adjacent.
• The remaining 10 tosses will be tails (T).

Step 2: Treat the tails (T) as separators

If there are 10 tails, then between and around these tails, there are possible positions to place the heads. We can imagine these tails as creating 11 "gaps":

• 1 gap before the first tail
• 1 gap between each pair of tails (9 gaps between tails)
• 1 gap after the last tail

For the heads not to be adjacent, each head must go into a different gap.

Step 3: Choose 5 gaps for the heads

We now need to choose 5 gaps from the 11 available gaps to place the 5 heads. The number of ways to choose 5 gaps from 11 is given by the combination formula:

$\binom{11}{5} = \frac{11!}{5!(11-5)!} = \frac{11 \times 10 \times 9 \times 8 \times 7}{5 \times 4 \times 3 \times 2 \times 1} = 462$

Conclusion:

There are 462 possible outcomes where exactly 5 heads appear, and no two heads are adjacent.

Would you like further explanation or have any questions?

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Here are 5 related questions to deepen understanding:

1. How many possible outcomes are there if no restrictions are placed on the arrangement of heads and tails?
2. What if the coin is tossed 10 times, and we need 3 heads with no two heads adjacent?
3. How many ways are there to arrange 6 heads and 9 tails without restrictions?
4. What is the probability of getting 5 heads in any sequence of 15 coin tosses?
5. How does the answer change if there must be at least one tail between any two heads?

Tip: For problems involving "no two items adjacent," think about placing the non-restricted items first (like tails here), and then count the available spaces to place the restricted items (like heads).