Question 1184450: A certain class has 11 students, 2 of whom are sisters. The class needs to be split across three rooms for a socially distanced exam, where room A can seat 2 students, room B 3 students, and room C 6 students. How many ways can the students be divided across the rooms, if the 2 sisters are not allowed to be in the same room (say, for academic integrity)?
Found 2 solutions by CPhill, ikleyn:
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! Here's how to solve this problem:
**1. Total Ways to Arrange Students (Without Restriction):**
First, let's calculate the total number of ways to arrange the students without considering the restriction about the sisters. We can think of this as choosing students for each room:
* Room A: Choose 2 students out of 11: ¹¹C₂ = (11!)/(2!9!) = 55 ways
* Room B: Choose 3 students from the remaining 9: ⁹C₃ = (9!)/(3!6!) = 84 ways
* Room C: The remaining 6 students go to room C: ⁶C₆ = 1 way
Multiply these together to get the total number of ways without restrictions: 55 * 84 * 1 = 4620
**2. Ways Sisters are Together:**
Now, let's calculate the number of ways the sisters *are* together and subtract this from the total. There are two scenarios where the sisters are together:
* **Sisters in Room A:**
* Choose the 2 sisters for Room A: 1 way
* Choose the remaining student for Room A: ⁹C₁ = 9 ways
* Choose 3 students for Room B: ⁸C₃ = 56 ways
* The remaining 6 students go to Room C.
* Total ways: 1 * 9 * 56 * 1 = 504 ways
* **Sisters in Room B:**
* Choose the 2 sisters for Room B: 1 way
* Choose the remaining student for Room B: ⁹C₁ = 9 ways
* Choose 2 students for Room A: ⁸C₂ = 28 ways
* The remaining 6 students go to Room C.
* Total ways: 1 * 9 * 28 * 1 = 252 ways
* **Sisters in Room C:**
* Choose the 2 sisters for Room C: 1 way
* Choose 4 more students for Room C: ⁹C₄ = 126 ways
* Choose 2 students for Room A: ⁵C₂ = 10 ways
* The remaining 3 students go to Room B.
* Total ways: 1 * 126 * 10 * 1 = 1260 ways
Add the ways the sisters can be together in each room: 504 + 252 + 1260 = 2016 ways.
**3. Ways Sisters are Separated:**
Subtract the number of ways the sisters are together from the total number of ways to arrange the students:
4620 - 2016 = 2604
Therefore, there are 2604 ways to divide the students such that the sisters are not in the same room.
Answer by ikleyn(52788)  (Show Source):
You can put this solution on YOUR website! .
A certain class has 11 students, 2 of whom are sisters. The class needs to be split across three rooms for a socially distanced exam,
where room A can seat 2 students, room B 3 students, and room C 6 students. How many ways can the students be divided across the rooms,
if the 2 sisters are not allowed to be in the same room (say, for academic integrity)?
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In the post by @CPhill, the strategy is good, but implementation has errors in calculations.
I came to fix and to correct his solution. My editing is right in the text.
Here's how to solve this problem:
**1. Total Ways to Arrange Students (Without Restriction):**
First, let's calculate the total number of ways to arrange the students without considering the restriction about the sisters. We can think of this as choosing students for each room:
* Room A: Choose 2 students out of 11: ¹¹C₂ = (11!)/(2!9!) = 55 ways
* Room B: Choose 3 students from the remaining 9: ⁹C₃ = (9!)/(3!6!) = 84 ways
* Room C: The remaining 6 students go to room C: ⁶C₆ = 1 way
Multiply these together to get the total number of ways without restrictions: 55 * 84 * 1 = 4620
**2. Ways Sisters are Together:**
Now, let's calculate the number of ways the sisters *are* together and subtract this from the total. There are two scenarios where the sisters are together:
* **Sisters in Room A:**
* Choose the 2 sisters for Room A: 1 way
* Choose the remaining student for Room A: ⁹C₁ = 9 ways <<<---=== error: there are no more places in A
* Choose 3 students for Room B: ⁸C₃ = 56 ways <<<---=== error: should be C(9,3) = 84
* The remaining 6 students go to Room C.
* Total ways: 1 * 9 * 56 * 1 = 504 ways <<<---=== error: should be 1*84 = 84.
* **Sisters in Room B:**
* Choose the 2 sisters for Room B: 1 way
* Choose the remaining student for Room B: ⁹C₁ = 9 ways
* Choose 2 students for Room A: ⁸C₂ = 28 ways
* The remaining 6 students go to Room C.
* Total ways: 1 * 9 * 28 * 1 = 252 ways
* **Sisters in Room C:**
* Choose the 2 sisters for Room C: 1 way
* Choose 4 more students for Room C: ⁹C₄ = 126 ways
* Choose 2 students for Room A: ⁵C₂ = 10 ways
* The remaining 3 students go to Room B.
* Total ways: 1 * 126 * 10 * 1 = 1260 ways
Add the ways the sisters can be together in each room: 504 + 252 + 1260 = 2016 ways. <<<---=== should be 84 + 252 + 1260 = 1596
**3. Ways Sisters are Separated:**
Subtract the number of ways the sisters are together from the total number of ways to arrange the students:
4620 - 2016 = 2604 <<<---=== should be 4620 - 1596 = 3024
Therefore, there are   ways to divide the students such that the sisters are not in the same room.
Fixed and corrected.
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