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This problem asks us to find the number of ways to get exactly 2 heads (H) and 4 tails (T) when flipping a balanced coin 6 times.

The scenario described follows a **binomial expansion** of \((H + T)^n\), where \(n\) is the number of flips, and each term gives the number of ways to get a particular combination of heads and tails.

To solve this, we use the **binomial coefficient** formula:
\[
\binom{n}{k} = \frac{n!}{k!(n-k)!}
\]
where:
- \(n = 6\) (total number of flips),
- \(k = 2\) (number of heads),
- \(n - k = 4\) (number of tails).

We need to calculate:
\[
\binom{6}{2} = \frac{6!}{2!(6-2)!} = \frac{6 \times 5}{2 \times 1} = 15
\]

Thus, there are **15 ways** to get exactly 2 heads and 4 tails when flipping the coin 6 times.

The correct answer is **B. 15**.

Would you like further details or have any questions?

Here are 5 related questions:
1. How would the binomial expansion look for 4 heads and 3 tails when flipping a coin 7 times?
2. What is the probability of getting exactly 3 heads when flipping a balanced coin 5 times?
3. How does the binomial coefficient change if we increase the number of flips from 6 to 8 but still want 2 heads?
4. How is the binomial theorem used in calculating different outcomes for dice rolls?
5. Can we apply a similar approach to find the number of combinations of outcomes when rolling two dice?

**Tip:** In binomial distributions, each term in the expansion corresponds to a specific combination of outcomes, and the coefficients represent the number of ways to achieve that combination.

A balanced coin with one side heads (H) and the other side tails (T) is repeatedly flipped, and the results are recorded. The coefficients in the expansion of (H + T)^n give the number of ways to get a particular combination of heads and tails for n flips of the coin. For example, in the expansion of (H + T)^7, the term 21H^2T^5 means there are 21 ways to get exactly 2 heads and 5 tails when flipping a coin 7 times. How many ways are there to get exactly 2 heads and 4 tails when flipping a balanced coin 6 times?

Learn how to calculate the number of ways to get exactly 2 heads and 4 tails when flipping a balanced coin 6 times. This problem uses the binomial coefficient and combinatorics. Suitable for Grades 10-12.

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