There are 6 girls and 4
boys running a race.  What is the probability that only girls will finish in the top three places and a boy in the last place?

To solve this problem, let's break it down step by step:

### 1. **Determine the total number of possible outcomes:**

Since there are 10 participants (6 girls and 4 boys), the total number of ways they can finish the race is the total permutations of 10 participants:

\[
\text{Total possible outcomes} = 10! = 10 \times 9 \times 8 \times \cdots \times 1
\]

### 2. **Determine the favorable outcomes:**

We need to count the number of favorable outcomes where:
- The top three positions are occupied by only girls.
- The last position is occupied by a boy.

#### Step 1: Selecting the top three positions for the girls.
- The number of ways to choose 3 girls from 6 is given by the combination \(\binom{6}{3}\).
- After choosing 3 girls, the number of ways to arrange these 3 girls in the top three positions is \(3!\).

\[
\text{Ways to arrange top 3 girls} = \binom{6}{3} \times 3! = \frac{6!}{(6-3)!3!} \times 6 = 20 \times 6 = 120
\]

#### Step 2: Choosing a boy to finish last.
- The number of ways to choose 1 boy from 4 to finish in the last place is \(\binom{4}{1}\), which is 4.

#### Step 3: Arranging the remaining 6 participants in the middle positions.
- There are 6 remaining participants (3 girls and 3 boys) left to fill the remaining 6 positions. The number of ways to arrange them is \(6!\).

\[
\text{Ways to arrange remaining 6 participants} = 6! = 720
\]

### 3. **Calculate the total number of favorable outcomes:**

Multiply the number of ways to arrange the top three girls, choose the boy for the last place, and arrange the remaining participants:

\[
\text{Favorable outcomes} = 120 \times 4 \times 720 = 345,600
\]

### 4. **Calculate the probability:**

The probability is the ratio of the number of favorable outcomes to the total possible outcomes:

\[
P(\text{only girls in top 3, boy in last place}) = \frac{345,600}{10!} = \frac{345,600}{3,628,800} = \frac{12}{126} = \frac{1}{105} \approx 0.0095
\]

So, the probability is approximately \(\frac{1}{105}\).

---

Do you want more details or have any questions?  
Here are 5 related questions to further your understanding:

1. What is the probability that all girls finish before all boys?
2. How would the probability change if there were 5 boys and 5 girls?
3. How many ways can the girls and boys be arranged if no restrictions are applied?
4. What is the probability that a specific girl wins the race?
5. What is the probability that girls finish in the top two positions?

**Tip:** Understanding combinations and permutations is crucial for solving probability problems like these.

Explore the probability calculation where 6 girls and 4 boys participate in a race. Understand how to compute the probability that only girls finish in the top three places with a boy in the last place using permutations and combinations. Suitable for high school students studying probability and permutations.

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