Four guests A, B, C and D have to be assigned into three different off

Question

Four guests A, B, C and D have to be assigned into three different office rooms (one double room and two single rooms) such that A and B won't have to stay in the double room. How many ways in total can they be assigned into these office rooms?

A. 5
B. 10
C. 12
D. 15
E. 20

Show Answer

Official Answer

B

A. 5
B. 10
C. 12
D. 15
E. 20

rsvelc · Answer

Number of ways to arrange 4 people i.e; A,B,C,D in one double room and two single rooms = 4C2 (Select 2 people who will go to one double room).
Number of ways to arrange A,B in one double room = 1

Total number of ways to arrange A,B,C,D in rooms so A and B don't endup in one double room = 4C2 - 1 = 5.

Let me know if I missed any considerations.

harshittahuja · Answer

Here we can say that
R1- Double room
R2,R3- Single rooms

__(R1),_(R2),_(R3)
in that A, B are not in R1
then simples C, D are in the R1 means that CD/DC can be two ways.
and A/B are in R1 and R2 two ways.
So we can say that answer is 2*2=4.

dream04 · Answer

We can as well use the slot method: AB _ _, AC _ _, AD _ _, BA _ _, BC _ _, BD _ _, CA _ _, CB _ _, CD _ _, DA _ _, DB _ _, DC _ _,.

The possibilities of A & B in the double room is 2 out of the 12 double. Hence, 12 - 2 = 10.

rsvelc · Answer

Hi  dream04  The above approach absolutely works if order arrangement is asked for.

For example: if we select CD and assign to the one double room it shouldn’t matter in what order they’re selected/arranged.

I really ask myself here - Does it really matter if I consider CD and DC as well? Is it not redundant if order is not what we’re looking for?

So the number of possibilities is reduced to 6(which is 4C2) 
Number of possibilities to select AB = 1

Total possibilities = 6-1 = 5

Let me know if you have any concerns with this approach

Posted from my mobile device

tmoot · Answer

There are 12 ways to arrange people into the two single rooms, 4 into the first and 3 into the second. This leaves the remaining two people in the double room (where order won't matter) so there is at most 4* 3 = 12 arrangements.

Considering only the single rooms, in the case that C & D are each assigned the single rooms then A & B are in the double- not good. There are 2 ways this could happen - C in the first double room and D in the second or vice versa.

So we get 12 - 2 = 10 possible arrangements.

Four guests A, B, C and D have to be assigned into three different off

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Four guests A, B, C and D have to be assigned into three different off

Prep Toolkit

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