3. A researcher has 14 subjects taking part in an experiment. They will randomly assign 5 subjects to the first group, 5 to the second, and 4 to the third. How many ways are there to choose which participants are assigned to each group? Leave your answer in exponential or factorial form.

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discrete math

**Problem:**

A researcher has 14 subjects taking part in an experiment. They will randomly assign 5 subjects to the first group, 5 to the second, and 4 to the third. How many ways are there to choose which participants are assigned to each group? *Leave your answer in exponential or factorial form.*

---

**Solution Explanation:**

To determine the number of ways to assign the subjects to the groups, we can use combinations. The total number of subjects is 14, and we need to decide how to divide them into groups of 5, 5, and 4.

1. **Choose 5 subjects for the first group:**
   \[
   \binom{14}{5}
   \]

2. **Choose 5 of the remaining 9 subjects for the second group:**
   \[
   \binom{9}{5}
   \]

3. **The last 4 subjects automatically go to the third group:**
   \[
   \binom{4}{4} = 1
   \]

The total number of ways to assign the subjects to these groups is the product of these combinations:
\[
\binom{14}{5} \times \binom{9}{5} \times \binom{4}{4}
\]

In factorial form, this is expressed as:
\[
\frac{14!}{5!5!4!}
\]
Transcribed Image Text:**Problem:** A researcher has 14 subjects taking part in an experiment. They will randomly assign 5 subjects to the first group, 5 to the second, and 4 to the third. How many ways are there to choose which participants are assigned to each group? *Leave your answer in exponential or factorial form.* --- **Solution Explanation:** To determine the number of ways to assign the subjects to the groups, we can use combinations. The total number of subjects is 14, and we need to decide how to divide them into groups of 5, 5, and 4. 1. **Choose 5 subjects for the first group:** \[ \binom{14}{5} \] 2. **Choose 5 of the remaining 9 subjects for the second group:** \[ \binom{9}{5} \] 3. **The last 4 subjects automatically go to the third group:** \[ \binom{4}{4} = 1 \] The total number of ways to assign the subjects to these groups is the product of these combinations: \[ \binom{14}{5} \times \binom{9}{5} \times \binom{4}{4} \] In factorial form, this is expressed as: \[ \frac{14!}{5!5!4!} \]
Expert Solution
Step 1

Total subjects = 14 

3 groups :

1: 5 subjects

2: 5 subjects

3: 4 subjects

No. of ways = ?

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