In how many ways can 5 boys be allotted four different rooms such that

Question

In how many ways can 5 boys be allotted four different rooms such that none of the rooms are empty and all the 5 boys are accommodated?

A) 5C2 * 4!
B) 5C3 * 5!
C) 5C4 * 4!
D) 5C1 * 3!
E) 5C1

Source: 4Gmat

krikatkat · Accepted Answer

Hi alphonsa,

I would try to explain

We have 4 rooms and 5 boys so necessarily 2 boys will stay in the same room.
First lets count the number of ways to pick the two boys- 5C2
Now we have 4 groups of boys - 1 boy, 1boy, 1 boy , 2boys that should be accommodated in 4 rooms - 4!

So the number of ways that 5 boys be allotted four different rooms is 5C2 *4!

I hope it helps :-D

GMATinsight · Answer

Because 4 rooms and 5 boys are available and each room must have atleast one boy so one Room must have two boys

Step 1: Select two boys who share one room in 5C2 ways (Now we will treat this group of two boys as one entity)

Step 2: Now we have 4 entities (including group of two boys as one entity) to arrange in 4 rooms which can be done in 4! ways

Total Ways of arranging boys in 4 rooms = 5C2*4! = 10*24 = 240

Answer: Option A

Ashishmathew01081987 · Answer

My method is as follows, please correct me if i am wrong

4 boys can be selected from 5 boys in 5C4 ways = 5
4 boys can be arranged in 4 rooms in 4! ways
and since 1 boy has to share a room with one of the 4 boys, therefore total ways = 2! = 2

Hence total number of ways in which allocation can be done = 5C4 * 4! * 2! = 5*2*4!

5*2 can be written as 5C2 so ans is A

Michael2016SZ · Answer

could somebody explain why 4 group to 4 rooms is 4! ?

sleepynut · Answer

Let me try.
At first,we have 5 boys and 4 rooms.
If none of the room is empty,it must be that one room accommodates 2 boys.
First,pick a room to accommodate these two boy.
Then,pick 2 out of 5 boys to fit in that room.
Now,we will have 3 rooms and 3 boys.The permutation tells us that there will be 3! ways to fit each of the remaining boys in the rooms.
Combining these steps together,we will have (4)(5C2)(3!) ways to place these boys under the given condition.

ganand · Answer

Hi Michael2016SZ,

For example, consider following:

Four rooms are Room no. 1,2,3, and 4.

Four groups are A, B, C, and D.

Room no. 1 can be filled by any one of the four group, i.e. either by A, B, C, or D (in 4 ways)
AND
Room No. 2 can be filled by any of the remaining three groups (in 3 ways)
AND
Room No. 3 can be now occupied by any of the remaining two groups (in 2ways)
AND
And finally, Room No. 4 will be filled by the last group. (in 1 way)

Total no. of ways = 4*3*2*1 = 4! (Always remember that AND => *, and OR => +) .

Hope this helps.

karek77 · Answer

Can you explain it to me why my way is wrong plz?!

Here is what I did:

You can arrange 5 boys in 4 rooms by 5*4*3*2 (5!) and you get 1 more boy to place in one of the 4 rooms so at the end you have 5!*4

pushpitkc · Answer

One of the rooms must have 2 boys, where as each of the other rooms can have 1 boy only.  
You can't just like that place the one student in either of the rooms.

The total ways to choose 2 boys in the first of the rooms is 5c2.
The other 3 rooms will have 3*2*1 ways of placing the remaining boys.
There are 4 ways of arranging these boys in the rooms, making the total number of arrangements : 10*3*2*4 = 240.

Hope that helps!

Vandanaa · Answer

@Bunuel@EgmatQuantExpert
 MartyMurray 
Choosing 4 people out of 5 = 5C4 = 5
Choosing the 4 rooms for these 4 people = 4! = 24
Choosing 1 among the 4 rooms for the remaining person = 4C1 = 4

So why is it not 5*24*4 = 480

Krunaal · Answer

We choose the 2 boys first to avoid double-counting, let's say by your logic we assign  A->R1, B->R2, C->R3, D->R4; now E is the 5th boy and he's allotted R1  .....(i)

Now, if suppose the 4 boys we select first were assigned  E->R1,   B->R2, C->R3, D->R4; now A is the 5th boy and he's allotted R1  .....(ii)

Above, (i) and (ii) led to the same case and were counted twice by your approach.

So we select the two boys first => 5C2, there are 4 to assign them into one of the rooms, and the remaining 3 boys can be arranged in 3! ways. Therefore, 5C2 * 4 * 3! =>  5C2 * 4!

In how many ways can 5 boys be allotted four different rooms such that

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Problem Solving (PS) Questions

In how many ways can 5 boys be allotted four different rooms such that

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards