2004 HSC Probablity Question!! (1 Viewer)

[kman16]

Hi,

I just attempted the 04 MX2 past paper and I am stumped on this probability question. Any help would be appreciated.

Q. In how many ways can "n" students be placed in two distinct rooms so that neither room is empty?

A. nC1 + nC2 + nC3 + ... + nC(n-1)

My answer was their answer multiplied by 2. Shouldn't you have to multiply each case by "2C1" as there are 2 DISTINCT rooms and each time you put one group in one room, you should have to choose the room you put them in?

Thanks

 

[Carrotsticks]

Consider the case when we place k people into 1 room.

First place the k people into that room, selected from a total of n people. So nCk.

Then place the remaining n-k people into the other room, selected from a total of n-k people. So (n-k)C(n-k).

There is no need to 'select a room' initially, because if we place some number of people into one room, that forces the remainder to go into the other room.

Since it forces them to the other room, this means that we have already, in a sense, 'selected' the room for them already by excluding them from the original room.

 

[Carrotsticks]

Here's a silly example, to get my idea across.

Suppose I'm a pirate (arrr) and I have 3 prisoners A B C. I have to select 2 of them to survive, and one of them to be doomed to the plank.

Suppose I have a grudge against A because he mocked my parrot.

I could specifically say "Yep, I don't like A, I want him to walk the plank"

Alternatively, I could say "Hmm, I like B and C, so I'll choose them to survive" (thereby dooming A).

Are these not the same options? The equivalent of you multiplying by 2, would be the equivalent of me treating "I want A to drown" and "I want B and C to survive" as different cases, when they are in fact the same.

 

[bobmcbob365]

So to make this a bit simpler, let the number of students 'n' be 3. Let the students be named A, B and C.

To keep this simple, I'll only consider the nC1 and nC(n-1) in the series.

So nC1 for Room 1 in our case would be 3C1 and the possible events would be

Room 1: A Room 2: B,C
Room 1: B Room 2: A,C
Room 1: C Room 2: A,B

For nC(n-1) in Room 1, in our case, it would be 3C2 and the possible events would be

Room 1: A,B Room 2: C
Room 1: A,C Room 2: B
Room 1: B,C Room 2: A

As you can see from each possible event from nC1 and nC(n-1), the latter is exactly the same as the former, but with the students in the other room. It is this that accounts for the different room that you "choose" the first person to be placed in. This is why the famous rule nCk = nC(n-k) applies.

Now the same thing happens if we take nC2 and nC(n-2) and so on.

The same thing also happens if we take more than 3 people.

 

[Trebla]

You only multiply if the events are independent. In this case, placing say k people in the 1st room automatically places the other n - k people in the 2nd room. So the event that people go into the 2nd room is in fact dependent on the number of people who in the 1st room due to the constraint of having exactly n people to allocate.

 

[seanieg89]

Equivalently:

How many ways can we put the n students into two rooms?

2^n. (there are two choices of room for each student.)


How many of these situations involve an empty room?

2. (the situations where they are all in room A and where they are all in room B.)


So the answer is: 2^n-2.

 

[kman16]

Here's a silly example, to get my idea across. Suppose I'm a pirate (arrr) and I have 3 prisoners A B C. I have to select 2 of them to survive, and one of them to be doomed to the plank.Suppose I have a grudge against A because he mocked my parrot.I could specifically say "Yep, I don't like A, I want him to walk the plank"Alternatively, I could say "Hmm, I like B and C, so I'll choose them to survive" (thereby dooming A).Are these not the same options? The equivalent of you multiplying by 2, would be the equivalent of me treating "I want A to drown" and "I want B and C to survive" as different cases, when they are in fact the same.

HAHA this made me laugh

Equivalently:How many ways can we put the n students into two rooms? 2^n. (there are two choices of room for each student.)How many of these situations involve an empty room?2. (the situations where they are all in room A and where they are all in room B.)So the answer is: 2^n-2.

This is a really clever way to look at it! I wish i had thought of it like that Thanks everyone for helping!