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)?
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~



        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.



<pre>
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 {{{highlight(cross(2604))}}}  {{{highlight(highlight(3024))}}}  ways to divide the students such that the sisters are not in the same room.
</pre>

Fixed and corrected.