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The question asks how many ways there are to get exactly 2 heads and 3 tails when a coin is tossed 5 times.

This problem can be solved using the binomial coefficient formula:
\[
\binom{n}{k} = \frac{n!}{k!(n-k)!}
\]
where:
- \(n\) is the total number of tosses,
- \(k\) is the number of heads (or tails), and
- \(\binom{n}{k}\) gives the number of ways to choose \(k\) heads from \(n\) tosses.

In this case:
- \(n = 5\) (since the coin is tossed 5 times),
- \(k = 2\) (since we want exactly 2 heads).

So, we need to compute:
\[
\binom{5}{2} = \frac{5!}{2!(5-2)!} = \frac{5 \times 4}{2 \times 1} = 10
\]

Thus, there are 10 ways to get exactly 2 heads and 3 tails.

The correct answer is **10**.

Do you want any further details or have any questions?

Here are 5 related questions:
1. What is the total number of possible outcomes when a coin is tossed 5 times?
2. How does the binomial expansion help in solving coin toss problems?
3. What if the problem asked for exactly 3 heads and 2 tails instead?
4. How would the solution change if the number of tosses increased to 7?
5. Can you explain the significance of the binomial theorem in probability?

**Tip:** Always break down a problem step by step to make it easier to solve!

If a coin is tossed 5 times, how many ways can the result be exactly 2 heads and 3 tails?

Learn how to solve a combinatorics problem by calculating the number of ways to get exactly 2 heads and 3 tails when tossing a coin 5 times. This solution uses the binomial coefficient formula, and is suitable for students in Grades 9-12.

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