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In how many ways can 5 boys be allotted four different rooms such that [#permalink]
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14 Sep 2014, 07:47
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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
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In how many ways can 5 boys be allotted four different rooms such that [#permalink]
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14 Sep 2014, 08:49
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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



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Re: In how many ways can 5 boys be allotted four different rooms such that [#permalink]
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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 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]
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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]
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16 Feb 2017, 00:21
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]
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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]
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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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Re: In how many ways can 5 boys be allotted four different rooms such that [#permalink]
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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]
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21 Feb 2017, 02:15
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
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Re: In how many ways can 5 boys be allotted four different rooms such that [#permalink]
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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]
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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]
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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]
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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]
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15 Jul 2017, 16:31
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]
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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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