The problem in the image involves determining the number of ways to get exactly 4 heads and 1 tail in 5 tosses of a balanced coin.

This is a combination problem. We can use the binomial coefficient formula, which gives the number of ways to choose 4 heads (or equivalently 1 tail) from 5 tosses. The formula is:

$\binom{n}{k} = \frac{n!}{k!(n-k)!}$

where $n = 5$ (the number of tosses) and $k = 4$ (the number of heads). Plugging in the values:

$\binom{5}{4} = \frac{5!}{4!(5-4)!} = \frac{5 \times 4!}{4! \times 1!} = \frac{5}{1} = 5$

Thus, there are 5 ways to get exactly 4 heads and 1 tail.

So, the correct answer is 5.

Would you like more details on the binomial coefficient? Let me know if you have any questions!

Here are 5 related questions to expand your understanding:

1. What is the total number of possible outcomes when tossing a coin 5 times?
2. How would you calculate the number of ways to get 3 heads and 2 tails in 5 tosses?
3. What would be the probability of getting exactly 4 heads in 5 tosses?
4. How does the binomial theorem relate to problems like these?
5. If you wanted to get at least 1 tail in 5 tosses, how many outcomes would satisfy that condition?

Tip: For problems involving combinations, think of it as "choosing" a specific number of successes (like heads) from a set number of trials (like coin tosses). The binomial coefficient helps in counting these possibilities.