A coin is flipped eleven times in succession. In how many ways can at least ten heads occur?

We need to determine how many ways at least 10 heads can occur when a coin is flipped 11 times. This means either 10 heads or 11 heads.

### Step 1: Calculate the number of ways to get exactly 10 heads
The number of ways to get exactly \( k \) heads in \( n \) flips is given by the binomial coefficient:
\[
\binom{n}{k}
\]
where \( n \) is the total number of flips and \( k \) is the number of heads.

For exactly 10 heads in 11 flips, we have:
\[
\binom{11}{10} = 11
\]
This is because choosing 10 heads is equivalent to choosing 1 tail (out of 11 flips), and there are 11 ways to do that.

### Step 2: Calculate the number of ways to get exactly 11 heads
For exactly 11 heads in 11 flips, there is only 1 way:
\[
\binom{11}{11} = 1
\]

### Step 3: Total number of ways
The total number of ways to get at least 10 heads is the sum of the two cases (10 heads and 11 heads):
\[
11 + 1 = 12
\]

Thus, there are \( \boxed{12} \) ways for at least 10 heads to occur when flipping a coin 11 times.

Would you like more details on how binomial coefficients work or have any other questions?

Here are 5 related questions for further practice:
1. How many ways can at least 9 heads occur in 11 flips?
2. What is the probability of getting exactly 10 heads in 11 flips?
3. How many ways can at least 2 tails occur in 11 flips?
4. How many ways can no heads occur in 11 flips?
5. How many ways can exactly 5 heads occur in 11 flips?

**Tip:** Binomial coefficients are a powerful tool for counting combinations and can be computed easily with the formula \( \binom{n}{k} = \frac{n!}{k!(n-k)!} \).

A coin is flipped eleven times in succession. In how many ways can at least ten heads occur?

Discover how to calculate the number of ways at least 10 heads can occur when a coin is flipped 11 times. This combinatorics problem covers the use of binomial coefficients, relevant to probability and counting principles. Suitable for students in Grades 9-12.

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