Individual Practice: 1. We need to use permutation to determine the number of ways 6 students can be seated in a row of 6 seats if 2 of the students insist on sitting beside each other. Is this correct? Why or why not? 2. A student has 9 friends, consisting of 4 boys and 5 girls. In how many ways can he invite them to his birthday party if he can only invite 3 boys and 3 girls?

Let's tackle each of your individual practice problems step by step. --- ### **1. Permutation for Seating 6 Students with 2 Students Sitting Together** **Problem Statement:** Determine the number of ways 6 students can be seated in a row of 6 seats if 2 of the students insist on sitting beside each other. **Is Using Permutation Correct?** Yes, using permutations is correct for solving this problem. Permutations are appropriate here because the order in which the students are seated matters, and we are arranging distinct objects (students) in a specific order. **Solution:** 1. **Treat the 2 Students as a Single Entity:** - Since the two students must sit together, we can consider them as a single "block." - This effectively reduces the problem to arranging **5 entities**: the 4 individual students + the 1 block of 2 students. 2. **Calculate the Arrangements of the 5 Entities:** - The number of ways to arrange 5 distinct entities in a row is given by \(5! = 120\) ways. 3. **Account for the Internal Arrangement of the 2 Students:** - Within the "block," the two students can switch places. There are \(2! = 2\) ways to arrange them. 4. **Total Number of Arrangements:** \[ \text{Total ways} = 5! \times 2! = 120 \times 2 = 240 \text{ ways} \] **Conclusion:** Using permutations is appropriate and correctly applied in this scenario, resulting in **240** possible seating arrangements where the two specific students are seated together. --- ### **2. Inviting Friends: Selecting 3 Boys and 3 Girls** **Problem Statement:** A student has 9 friends, consisting of 4 boys and 5 girls. In how many ways can he invite them to his birthday party if he can only invite **3 boys and 3 girls**? **Solution:** This is a combination problem since the order in which the friends are invited does not matter—only the selection of which friends to invite. 1. **Selecting 3 Boys from 4:** - The number of ways to choose 3 boys out of 4 is given by the combination formula: \[ \binom{4}{3} = 4 \text{ ways} \] 2. **Selecting 3 Girls from 5:** - The number of ways to choose 3 girls out of 5 is: \[ \binom{5}{3} = 10 \text{ ways} \] 3. **Total Number of Ways to Invite 3 Boys and 3 Girls:** - Since the selections are independent, multiply the number of ways to choose the boys by the number of ways to choose the girls: \[ \text{Total ways} = \binom{4}{3} \times \binom{5}{3} = 4 \times 10 = 40 \text{ ways} \] **Conclusion:** There are **40** different ways for the student to invite 3 boys and 3 girls from his group of friends. --- Feel free to ask if you have any further questions or need additional clarifications!