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Dividing 200 people in 3 unequal rooms and probability of certain combinations

 
Mathematics
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Topic starter 19/03/2023 5:36 pm  

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We have 3 rooms (say 1, 2 and 3). Room 1 can take 50 people, room 2 can take 50 and room 3 can take 100. I understand that the number of ways of assigning 200 people to these 3 rooms is $200 choose 50$$ 150 choose 50$. The question is what is the probability that Alice (A) and Bob (B) end up in the same room while Charlotte (C) ends up in a different room.

My solution

I first divided the event E = {A and B in same room, C in different room} in the disjoint events E1 = {A and B in room 1, C in room 2} $cup$ {A and B in room 1, C in room 3} , E2 = {A and B in room 2, C in room 1} $cup$ {A and B in room 2, C in room 3} and E3 = {A and B in room 3, C in room 1} $cup$ {A and B in room 3, C in room 2}.

I calculated the possible cases for E1 in the following way: $197 choose 48$$149 choose 49$
$+$ $197 choose 47$ $149 choose 50$. My reasoning for the case A and B in 1 while C in 2 was that I first pick the people that will make up room 1 alongside A and B (excluding C) and then pick the people that will make up room 2 alongside C (room 3 will then be uniquely determined). The reasoning for the case A and B in 1 while C in 3 was analogous.

Similar reasonings were employed to calculate the favorable cases for E2 and E3:

$|E2|$ = $197 choose 49$$148 choose 48$
$+$ $197 choose 50$ $147 choose 48$

$|E3|$ = $197 choose 49$$148 choose 50$
$+$ $197 choose 50$ $147 choose 49$

The probability was then simply calculated by summing the favorable cases for E1, E2 and E3 and finally dividing by the total number of cases.

I am wondering if this method is correct. Furthermore, if it is correct, I am wondering if there is a simpler method since this method requires dealing with awfully big numbers.

Thank you for your help.

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