2004 HSC - Q5(b)(ii) (1 Viewer)

[Stefano]

In how many ways can five students be placed in three distinct rooms so that no room is empty?

I don't know how to approach these questions. Is there any theory behind it or just logic?
All I know is that the answer involves a 3^5 somewhere. Can anyone help me fill in the gaps ?

 

[Slidey]

Could you do part i? One method of solving this problem is to use the method in part i.

 

[KFunk]

Starting with part i) the number of ways to place n students into 2 rooms so that neither room is empty = (2ⁿ - 2) since there are 2 options for each student and you have to subtract the two cases where the first or the second room is empty.

For part ii) it may well be helpful to use the logic of part i), as Slide said.

Using the same logic as in i) you know that each student can be put in one of three rooms yielding 3⁵ possible arrangments but you then need to subtract the options which include empty rooms. You can easily work out the number of options where 2 rooms are empty. If you look at options where one of the three rooms is empty you will realise that you're looking at distributing the 5 students between 2 rooms... looks kinda like part i). Give it a go.

# ways = 3⁵ - (ways with 2 rooms empty) - (ways with 1 room empty)

 

[KFunk]

Is there any theory behind it or just logic?

If there's any theory I geuss it's that the number of ways to arrange p objects amongst q rooms (where each object and room is distinct) is q^p. The includes options where rooms are empty. The logic of this can be understood pretty easily from the example in i) where you have n objects, each with 2 options. For each object you multiply by 2 giving you 2ⁿ.

 

[Stefano]

Starting with part i) the number of ways to place n students into 2 rooms so that neither room is empty = (2ⁿ - 2) since there are 2 options for each student and you have to subtract the two cases where the first or the second room is empty.

For part ii) it may well be helpful to use the logic of part i), as Slide said.

Using the same logic as in i) you know that each student can be put in one of three rooms yielding 3⁵ possible arrangments but you then need to subtract the options which include empty rooms. You can easily work out the number of options where 2 rooms are empty. If you look at options where one of the three rooms is empty you will realise that you're looking at distributing the 5 students between 2 rooms... looks kinda like part i). Give it a go.

# ways = 3⁵ - (ways with 2 rooms empty) - (ways with 1 room empty)

i) why is there a '-2' in the solution?

ii) # ways = 3⁵ - 1⁵ - 3[2⁵]
= 146
= Incorrect (Where did I go wrong?)

 

[KFunk]

i) why is there a '-2' in the solution?

ii) # ways = 3⁵ - 1⁵ - 3[2⁵]
= 146
= Incorrect (Where did I go wrong?)

i) For you first question I'm sure you will agree that there are 2ⁿ ways to arrange n people between two rooms right? What you need to realise is that in one of these situations *everyone* has been put into room 1 and that in another situation *everyone* has been put into room 2. Since the questions asks for the number of arrangments where neither room is empty you have to subtract two from the total number of ways giving you: (2ⁿ - 2)

 

[KFunk]

ii) # ways = 3⁵ - 1⁵ - 3[2⁵]
= 146
= Incorrect (Where did I go wrong?)

First of all there are three options where 2 rooms are empty. You can put all the people in room 1, all the people in room 2 or all the people in room 3. This gives you '3' rather than 1⁵.

If only one room is to be empty you are left with two rooms to distribute the people between. From part i) we know that there are (2⁵ - 2) ways of doing this. We also subtract two here because if we didn't we would be including a case where 2 rooms are empty and we have already subtracted that above. There are 3 three individual rooms which could be left empty so the number of situations with one room empty is 3(2⁵ - 2). This gives us a net sum of:

# ways = 3⁵ - 3 - 3(2⁵ - 2) = 150

Edit: My spelling sucks wang.

 

[Stefano]

Thank you KFunk!

Well explained.

 

[KFunk]

No problem