In how many ways can 5 boys be allotted four different rooms such that : Problem Solving (PS)

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Ashishmathew01081987

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Re: In how many ways can 5 boys be allotted four different rooms such that [#permalink]

14 Sep 2014, 08:59

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krikatkat wrote:

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

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

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Re: In how many ways can 5 boys be allotted four different rooms such that [#permalink]

15 Feb 2017, 23:49

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

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In how many ways can 5 boys be allotted four different rooms such that [#permalink]

16 Feb 2017, 00:21

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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.

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Re: In how many ways can 5 boys be allotted four different rooms such that [#permalink]

16 Feb 2017, 00:24

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Michael2016SZ wrote:

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

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.

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Re: In how many ways can 5 boys be allotted four different rooms such that [#permalink]

21 Feb 2017, 01:20

ganand wrote:

Michael2016SZ wrote:

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

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.

it's clear for me , thank you!

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Michael2016SZ

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Re: In how many ways can 5 boys be allotted four different rooms such that [#permalink]

21 Feb 2017, 01:21

ganand wrote:

Michael2016SZ wrote:

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

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.

it does hlep , thank you !

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Re: In how many ways can 5 boys be allotted four different rooms such that [#permalink]

28 Jun 2017, 01:54

5 Boys in 4 Rooms with no room vacant can happen only if - 3 Rooms have 1 boy & 1 Room has 2 boys
We have to select & arrange - Consider the Arrangement in 4 rooms 1 1 1 2 = 4!/3! = 4 Arrangements possible
Now lets select the boys for the rooms -
5C1*4C1*3C1*2C2= 60
Therefore total nof ways = 60*4=240
Hence the answer 5C2*4!

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Re: In how many ways can 5 boys be allotted four different rooms such that [#permalink]

15 Jul 2017, 13:45

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

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Re: In how many ways can 5 boys be allotted four different rooms such that [#permalink]

15 Jul 2017, 14:03

karek77 wrote:

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

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!

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Re: In how many ways can 5 boys be allotted four different rooms such that [#permalink]

15 Jul 2017, 14:08

pushpitkc wrote:

karek77 wrote:

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

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!

I understand why 5c2*4! is right. But i dont know why what I did is wrong

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Re: In how many ways can 5 boys be allotted four different rooms such that [#permalink]

15 Jul 2017, 16:31

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Please, correct me if I am wrong but the answer is correct because the boys are distinct.
I mean, if it were 5 apples in 4 baskets, the answer would be 5C2 * 4! / 3!, wouldn't it??
Thanks!

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In how many ways can 5 boys be allotted four different rooms such that [#permalink]

07 Jan 2018, 09:23

GMATinsight wrote:

alphonsa wrote:

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

Please explain

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

GMATinsight
hi

I have got your point very right, but
can you please tell me why the below process is not workable here?

total number of arrangement in which there is no restriction

= 4^5

now, number of ways in which 3 rooms are empty
+
number of ways in which 1 room is empty
+
number of ways in which 2 rooms are empty

now, if we subtract all these unacceptables from 4^5, what happens ??

please do say to me and please help me understanding the full concept
????

thanks in advance

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Re: In how many ways can 5 boys be allotted four different rooms such that [#permalink]

13 Mar 2019, 17:32

I understand the answer , but don't know why I getting wrong :
can any one explain ?

4 rooms , 5 boys ,
let's choose 4 out of the 5 boys and put each in a room : 5c4 * 4!
one boy left , he has 4 rooms to choose , so , the total # : 5c4 * 4! *4 = 5 * 4 * 4!
not 5 *2 * 4! as the answer here ! why ???

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In how many ways can 5 boys be allotted four different rooms such that [#permalink]

14 Mar 2019, 00:21

chetan2u Bunuel generis
I agree with the answer. However I have a question of my own.
If I solve it this way:
Lets first allot 4 boys to each of the 4 rooms so that no room goes empty..We can select 4 boys in 5C4 ways and they can be allotted in the 4 rooms in 4! ways.Now we have one boy remaining.He can be allotted to any of the 4 rooms in 4 ways...So ans is 4*5C4*4!...
Please explain where did I go wrong...
Thanks in advance

P

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Re: In how many ways can 5 boys be allotted four different rooms such that [#permalink]

14 Mar 2019, 07:38

alphonsa wrote:

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

Please explain

Hi,
Given set is of 5 boys and the numbers of rooms is 4. It means there will be two boys in any one room.
Case 1. So the number of combinations we can form a group of two boys =5C2

Case 2. Now, the question becomes a permutation.
The number of ways we can fill 1st room = 4
The number of ways we can fill 2nd room = 3
The number of ways we can fill 3rd room = 2
The number of ways we can fill 4th room = 1

Or just 4!

So, the total ways will be Case 1 AND (*) Case 2 = 5C2 * 4! Option A

Cheers!!

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Re: In how many ways can 5 boys be allotted four different rooms such that [#permalink]

26 Jun 2020, 06:17

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Ashishmathew01081987 wrote:

krikatkat wrote:

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

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

One last boy can choose among 4 rooms, thus 4 ways. Not 2.
am I missing something?

B

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Re: In how many ways can 5 boys be allotted four different rooms such that [#permalink]

19 Apr 2021, 08:23

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GMATinsight wrote:

alphonsa wrote:

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

Please explain

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

If the question was interchanged like below, would we have a different answer?

In how many ways can 4 rooms be allotted 5 boys such that none of the rooms are empty and all the 5 boys are accommodated?

gmatclubot

Re: In how many ways can 5 boys be allotted four different rooms such that [#permalink]

19 Apr 2021, 08:23

GMAT Problem Solving (PS) Questions

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