Last visit was: 22 Apr 2026, 22:22 It is currently 22 Apr 2026, 22:22
Close
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Close
Request Expert Reply
Confirm Cancel
avatar
rpm
Joined: 21 Apr 2010
Last visit: 20 Jun 2014
Posts: 7
Own Kudos:
103
 [100]
Given Kudos: 2
Posts: 7
Kudos: 103
 [100]
6
Kudos
Add Kudos
92
Bookmarks
Bookmark this Post
Most Helpful Reply
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 22 Apr 2026
Posts: 109,763
Own Kudos:
Given Kudos: 105,850
Products:
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 109,763
Kudos: 810,696
 [51]
28
Kudos
Add Kudos
23
Bookmarks
Bookmark this Post
User avatar
KarishmaB
Joined: 16 Oct 2010
Last visit: 21 Apr 2026
Posts: 16,439
Own Kudos:
79,389
 [12]
Given Kudos: 484
Location: Pune, India
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 16,439
Kudos: 79,389
 [12]
9
Kudos
Add Kudos
3
Bookmarks
Bookmark this Post
General Discussion
User avatar
iambroke
Joined: 24 Mar 2010
Last visit: 05 May 2014
Posts: 68
Own Kudos:
39
 [6]
Given Kudos: 12
Posts: 68
Kudos: 39
 [6]
6
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Another way:

Youngest niece needs to choose between S,S, E & T.
case 1) chooses S ==> 4! ways to distribute rest of toys
case 2) doesnt choose S ==> 2 ways * (distribute SSXX to 4 children) = 2 * 4!/2! = 24

Total = 24+24 = 48...
User avatar
GMATD11
Joined: 10 Nov 2010
Last visit: 25 Sep 2012
Posts: 128
Own Kudos:
Given Kudos: 22
Location: India
Concentration: Strategy, Operations
GMAT 1: 520 Q42 V19
GMAT 2: 540 Q44 V21
WE:Information Technology (Computer Software)
GMAT 2: 540 Q44 V21
Posts: 128
Kudos: 3,328
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Pls chk the image

as per my solution

youngest gal can take doll from 5dolls except one.other gal can also take from rest 4 nd so on

4*4*3*2*1=96

But OA is 48
Attachments

gettting of a doll.JPG
gettting of a doll.JPG [ 29.22 KiB | Viewed 39127 times ]

User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 22 Apr 2026
Posts: 109,763
Own Kudos:
810,696
 [7]
Given Kudos: 105,850
Products:
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 109,763
Kudos: 810,696
 [7]
4
Kudos
Add Kudos
3
Bookmarks
Bookmark this Post
Gordon buys 5 dolls for his 5 nieces. The gifts include 2 identical Sun-and-Fun beach dolls, one Elegant Eddie dress-up doll, one G.I. Josie army doll, and one Tulip Troll doll. If the youngest niece doesn't want the G.I. Josie doll, in how many different ways can he give the gifts?

5 nieces: 1 - 2 - 3 - 4- 5
5 dolls: - S - S - E - G- T

1 doesn't want G.

Now if she gets E then the other four dolls (SSGT) can be assigned in 4!/2! ways (permutation of 4 letters out of which 2 S's are identical), the same if she gets T, and if gets S then the other four dolls (SEGT) can be assigned in 4! ways: 4!/2!+4!/2!+4!=48.

Or total was to assign SSEGT to 5 nieces is 5!/2! and ways to assign G to 1 is 4!/2! (the same as E to 1), so desired=total-restriction=5!/2!-4!/2!=48.
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 22 Apr 2026
Posts: 109,763
Own Kudos:
810,696
 [1]
Given Kudos: 105,850
Products:
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 109,763
Kudos: 810,696
 [1]
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
subhashghosh
Hi Bunuel

I calculated as :

# of ways a doll can be given to the youngest niece = 3 (as 2 are identical in 4)

Then the remaning # of dolls can be distributed in 4!/2! ways.

So total # of ways = 4!/2! * 3 = 36.

Could you please tell me where I'm wrong, i.e, why considering "3" is wrong ?

Regards,
Subhash

If she gets E or T then yes ways to distribute other 4 dolls will be 4!/2!, but if gets she gets S then the other four dolls (SEGT) can be distributed in 4! ways not 4!/2! as all 4 dolls in this case are distinct. So the answer is 2*4!/2!+4!=48 not 3*4!/2!=36.

Hope it's clear.
User avatar
144144
Joined: 08 Nov 2010
Last visit: 26 May 2013
Posts: 192
Own Kudos:
Given Kudos: 161
GPA: 3.9
WE 1: Business Development
Posts: 192
Kudos: 560
Kudos
Add Kudos
Bookmarks
Bookmark this Post
...and ways to assign G to 1 is 4!/2! (the same as E to 1)...

Eh, i understand why its 4!/2! when its 4 dolls to 2 girls when 2 is identical.

buy why its the same if u give her G? can u please explain?

thanks.
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 22 Apr 2026
Posts: 109,763
Own Kudos:
810,696
 [1]
Given Kudos: 105,850
Products:
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 109,763
Kudos: 810,696
 [1]
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
144144
...and ways to assign G to 1 is 4!/2! (the same as E to 1)...

Eh, i understand why its 4!/2! when its 4 dolls to 2 girls when 2 is identical.

buy why its the same if u give her G? can u please explain?

thanks.

When you give one doll to the youngest niece you you are left with 4 dolls to assign to 4 sisters. If you give the youngest niece E, G or T then 4 dolls left will have 2 identical S's and # of ways to distribute will be 4!/2! and if you give the youngest niece S then all 4 dolls left will be distinct so # of ways to distribute them will be 4!.

So what's the difference whether you give the youngest niece E or G? In both cases you distribute 4 out which 2 are identical.
User avatar
verycoolguy33
Joined: 28 Dec 2010
Last visit: 02 Apr 2012
Posts: 17
Own Kudos:
Given Kudos: 2
Posts: 17
Kudos: 41
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Hi Bunel ,
To your first post :
Total # of ways to distribute SSEGT among 5 sisters (without restriction) is !5/!2 =60 ;I am trying to understand how did you came to this !5/!2 ?

Is it a permutation of picking 5 out of 5 where 2 are same - 5P2/!2 ?
If this is correct so can it be like if there were 3 sisters instead of 5 , with all other condition intact ,the solution would have been -

Total # of ways to distribute SSEGT among 3 sisters (without restriction) is 5P3/!2 = 15;
The # of ways when the youngest niece gets G is: 4P2/!2 = 6 (give G to youngest and then distribute SSET among 2 sisters).

So, # of ways when youngest niece doesn't get G is:15-6 = 9 .

Please explain for better understanding . Thanks.
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 22 Apr 2026
Posts: 109,763
Own Kudos:
Given Kudos: 105,850
Products:
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 109,763
Kudos: 810,696
Kudos
Add Kudos
Bookmarks
Bookmark this Post
verycoolguy33
Hi Bunel ,
To your first post :
Total # of ways to distribute SSEGT among 5 sisters (without restriction) is !5/!2 =60 ;I am trying to understand how did you came to this !5/!2 ?

Is it a permutation of picking 5 out of 5 where 2 are same - 5P2/!2 ?
If this is correct so can it be like if there were 3 sisters instead of 5 , with all other condition intact ,the solution would have been -

Total # of ways to distribute SSEGT among 3 sisters (without restriction) is 5P3/!2 = 15;
The # of ways when the youngest niece gets G is: 4P2/!2 = 6 (give G to youngest and then distribute SSET among 2 sisters).

So, # of ways when youngest niece doesn't get G is:15-6 = 9 .

Please explain for better understanding . Thanks.

I'm not sure understood your question completely. Below is explanation for 5!/2!.

1 - 2 - 3 - 4- 5 (5 nieces)
S - S - E - G - T (5 dolls)
S - E - S - G - T
E- S - S - G - T
...

So as you can see # ways to assign these dolls to five sisters is basically # of permutations of 5 letters SSEGT.

THEORY.
Permutations of \(n\) things of which \(P_1\) are alike of one kind, \(P_2\) are alike of second kind, \(P_3\) are alike of third kind ... \(P_r\) are alike of \(r_{th}\) kind such that: \(P_1+P_2+P_3+..+P_r=n\) is:

\(\frac{n!}{P_1!*P_2!*P_3!*...*P_r!}\).

For example number of permutation of the letters of the word "gmatclub" is \(8!\) as there are 8 DISTINCT letters in this word.

Number of permutation of the letters of the word "google" is \(\frac{6!}{2!2!}\), as there are 6 letters out of which "g" and "o" are represented twice.

Number of permutation of 9 balls out of which 4 are red, 3 green and 2 blue, would be \(\frac{9!}{4!3!2!}\).

So, for our case # of permutations of 4 letters SSEGT out of which 2 S's are identica is \(\frac{5!}{2!}\).

Hope it's clear.
User avatar
verycoolguy33
Joined: 28 Dec 2010
Last visit: 02 Apr 2012
Posts: 17
Own Kudos:
Given Kudos: 2
Posts: 17
Kudos: 41
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Hi Bunuel - Thanks for your reply , but this does not answer my qs. Let me re-phrase my qs .
In the original doll-nieces problem , if there were 3 nieces instead of 5 , so rephrasing the problem -
Q: Gordon buys 5 dolls for his 3 nieces. The gifts include 2 identical "S" dolls, one "E" doll, one "J" doll and one "T" doll. If the youngest niece does not want the "J" doll, in how many different ways can he give the gifts?
my Ans :
Total # of ways to distribute SSEGT among 3 sisters (without restriction) is 5P3/!2 = 15;
The # of ways when the youngest niece gets G is: 4P2/!2 = 6 (give G to youngest and then distribute SSET among 2 sisters).
So, # of ways when youngest niece doesn't get G is:15-6 = 9

Please explain .
Thanks,
User avatar
KarishmaB
Joined: 16 Oct 2010
Last visit: 21 Apr 2026
Posts: 16,439
Own Kudos:
Given Kudos: 484
Location: Pune, India
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 16,439
Kudos: 79,389
Kudos
Add Kudos
Bookmarks
Bookmark this Post
verycoolguy33
Hi Bunuel - Thanks for your reply , but this does not answer my qs. Let me re-phrase my qs .
In the original doll-nieces problem , if there were 3 nieces instead of 5 , so rephrasing the problem -
Q: Gordon buys 5 dolls for his 3 nieces. The gifts include 2 identical "S" dolls, one "E" doll, one "J" doll and one "T" doll. If the youngest niece does not want the "J" doll, in how many different ways can he give the gifts?
my Ans :
Total # of ways to distribute SSEGT among 3 sisters (without restriction) is 5P3/!2 = 15;
The # of ways when the youngest niece gets G is: 4P2/!2 = 6 (give G to youngest and then distribute SSET among 2 sisters).
So, # of ways when youngest niece doesn't get G is:15-6 = 9

Please explain .
Thanks,

No, this is not correct. You need to be extremely careful in P&C. One tiny change can change the whole question.
If 5 dolls (S, S, E, G , T) have to be distributed among 3 nieces, how can you distribute them? It depends on which 3 dolls you are distributing. Say you pick S, E, T, there are 3! ways but if you pick S, E, G, you have fewer ways because G cannot go to the youngest niece. So you need to take cases.

Case 1: All dolls distinct. G not included.
Pick 3 of the 4 distinct dolls such that G is not included. You can do it in 1 way: S, E, T
Distribute these 3 among the 3 sisters in 3! ways

Case 2: All dolls distinct. G included.
Pick G and any two of the remaining 3 dolls in 3C2 i.e. 3 ways.
G can be distributed in 2 ways and the rest of the 2 dolls in 2! ways

Case 3: 2 dolls identical and another (not G)
The two identical dolls must be S, S. The unique doll can be chosen in 2 ways.
The unique doll can be distributed in 3 ways and (S, S) will be given to the other 2 sisters in only 1 way since both the dolls are identical.

Case 4: 2 dolls identical and G
The two identical dolls must be S, S.
G can be distributed in 2 ways (to any of the 2 elder nieces)
Then (S, S) can be distributed in only one way.

Total number of ways = 3! + 3*2*2! + 2*3*1 + 2*1 = 26

or do it the reverse way.

Calculate total number of ways of distributing the dolls among 3 nieces and subtract the number of ways in which youngest niece gets G
Total number of ways:
Case 1: All dolls distinct
Select 3 of the 4 distinct dolls in 4C3 ways = 4.
Distribute them among 3 sisters in 3! ways
Case 2: Two dolls identical, one unique
Select S, S and a unique doll in 3 ways.
Give the unique doll in 3 ways (to any of the 3 sisters) and then distribute the identical dolls in 1 way
Total number of ways = 4*3! + 3*3 = 33

Number of ways in which the youngest gets G:
Give doll G to youngest in 1 way.
Now there are 4 dolls {S, S, E, T} and 2 nieces
Case 1: Give distinct dolls
Select 2 out of 3 distinct dolls in 3C2 = 3 ways
Distribute the 2 dolls to the 2 nieces in 2! ways
Case 2: Give identical dolls
Give S, S to the two nieces in 1 way.

Total number of ways in which youngest gets G = 3*2! + 1 = 7

Number of ways in which the youngest doesn't get G = 33 - 7 = 26
User avatar
verycoolguy33
Joined: 28 Dec 2010
Last visit: 02 Apr 2012
Posts: 17
Own Kudos:
Given Kudos: 2
Posts: 17
Kudos: 41
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Thanks Karishma , I find the 2nd approach quite easy to understand , but fear that during actual test this could hardly be completed within 2-3 min time without a silly mistake !

Thanks a lot .
User avatar
KarishmaB
Joined: 16 Oct 2010
Last visit: 21 Apr 2026
Posts: 16,439
Own Kudos:
79,389
 [2]
Given Kudos: 484
Location: Pune, India
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 16,439
Kudos: 79,389
 [2]
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
verycoolguy33
Thanks Karishma , I find the 2nd approach quite easy to understand , but fear that during actual test this could hardly be completed within 2-3 min time without a silly mistake !

Thanks a lot .

Yes, you are right. The question isn't very hard but the cases you need to take get to you during the test. But I would not expect a similar question in the test. It doesn't have the zing which high level GMAT questions have. The question just involves making case after case methodically. GMAT questions don't usually involve too many steps if you know your concepts well.
User avatar
Sabby
Joined: 17 Jun 2018
Last visit: 04 Mar 2026
Posts: 84
Own Kudos:
Given Kudos: 397
Location: France
GPA: 3.85
Posts: 84
Kudos: 118
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Hi,

Way to distribute 5 dolls with 2 identical ones : 5!/2! = 60 ways
Ways where the youngest niece gets the doll G is by already assigning it to her and counting the ways others get the rest of the dolls : 4!/2!= 12
60-12=48
Answer D)
User avatar
RahulJain293
Joined: 24 Apr 2022
Last visit: 25 May 2025
Posts: 166
Own Kudos:
Given Kudos: 96
Location: India
Concentration: General Management, Nonprofit
GMAT Focus 1: 585 Q81 V80 DI76
GMAT Focus 1: 585 Q81 V80 DI76
Posts: 166
Kudos: 103
Kudos
Add Kudos
Bookmarks
Bookmark this Post
GMATD11
Pls chk the image

as per my solution

youngest gal can take doll from 5dolls except one.other gal can also take from rest 4 nd so on

4*4*3*2*1=96

But OA is 48

Bunuel I tried this way

All possible ways of assigning 5 people = 5! which is 60.

Now the questions asks us that G is not to be assigned to a particular person... So I tried the complement way (if I can call it that)

Not getting G to a particular niece = total ways - ways of getting G to a niece

Ways of getting G to a niece - G given to youngest leaving us with assigning 4 to other 4 nieces
which is 4! / 2! = 12

Hence NOT GETTING G = 60-12 = 48
User avatar
sayan640
Joined: 29 Oct 2015
Last visit: 22 Apr 2026
Posts: 1,120
Own Kudos:
861
 [1]
Given Kudos: 789
GMAT 1: 570 Q42 V28
Products:
GMAT 1: 570 Q42 V28
Posts: 1,120
Kudos: 861
 [1]
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Please check the highlighted part of your explanation. Can we say that this is actually  2  c 2 = 1 way ( for distributing 2 idential dolls among 2 girls ) ?
KarishmaB
Note here that the youngest niece can be assigned a doll in only 3 ways: One of the S dolls (they are both identical so it doesn't matter which one she gets) or E doll or T doll
How you would assign the rest of the dolls would depend on which doll the youngest one got. If she got an S doll, you can assign a doll to the next niece in 4 ways: S or E or G or T. If she got, say, the E doll, you assign a doll to the next niece in 3 ways: S or G or T.
This complicates this method.

Instead, try and assign nieces to the dolls since all nieces are distinct.
G doll can be assigned a niece in 4 ways (the youngest doesn't want her)
E doll can be assigned a niece in 4 ways again (the remaining 4 after one niece has been assigned to G doll)
T doll can be assigned a niece in 3 ways (remaining 3 nieces)
Now we have 2 identical dolls and 2 nieces. How will you assign them? You will give the nieces 1 doll each. There is no other way. Both dolls are same so it doesn't matter who gets which one.

Total number of allocations = 4*4*3 = 48

(or use one of the other great methods discussed above)
­
avatar
ManifestDreamMBA
Joined: 17 Sep 2024
Last visit: 21 Feb 2026
Posts: 1,387
Own Kudos:
Given Kudos: 243
Posts: 1,387
Kudos: 897
Kudos
Add Kudos
Bookmarks
Bookmark this Post
The dolls to be distributed with no restrictions = 5!/2! = 60
If youngest one had doll G, the rest of the dolls can be distributed = 4!/2! = 12

The ways dolls can be gifted given the constraints = 60 - 12 = 48
rpm
Gordon buys 5 dolls for his 5 nieces. The gifts include two identical S beach dolls, one E, one G, one T doll. If the youngest niece doesn't want the G doll, in how many different ways can he give the gifts?

A. 12
B. 24
C. 36
D. 48
E. 60

My ans: Total no. of ways to give gifts = 5P5 /2! = 5!/2! = 60.
How do I account for the youngest niece's condition?
User avatar
luisdicampo
Joined: 10 Feb 2025
Last visit: 19 Apr 2026
Posts: 480
Own Kudos:
Given Kudos: 328
Products:
Posts: 480
Kudos: 73
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Deconstructing the Question

We distribute 5 dolls to 5 nieces.
Dolls: S, S, E, G, T (two identical S).
The youngest niece cannot receive G.

First count all arrangements, then subtract the invalid ones.

Step-by-step

Total arrangements with two identical S:
\(5!/2! = 120/2 = 60\)

Invalid cases (youngest gets G):
Distribute remaining \(S, S, E, T\) to 4 nieces:
\(4!/2! = 24/2 = 12\)

Valid arrangements:
\(60 - 12 = 48\)

Answer: D (48)
 1   2   
Moderators:
Math Expert
109763 posts
Tuck School Moderator
853 posts