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# Gordon buys 5 dolls for his 5 nieces. The gifts include two

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Joined: 02 Sep 2009
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20 Feb 2012, 02:02
verycoolguy33 wrote:
Hi Bunel ,
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.
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Joined: 28 Dec 2010
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21 Feb 2012, 09:01
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

Thanks,
Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 8678
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23 Feb 2012, 20:22
verycoolguy33 wrote:
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

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
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Karishma
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25 Feb 2012, 06:28
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 .
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Posts: 8678
Location: Pune, India

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27 Feb 2012, 04:01
1
verycoolguy33 wrote:
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.
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Karishma
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GMAT 1: 640 Q48 V31
GPA: 3.45
WE: Marketing (Pharmaceuticals and Biotech)

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27 Feb 2012, 07:03
Applying the similar logic the number of ways when youngest niece gets G doll is 12 ..
hence total number of times when youngest does not get G is 60-12 = 48
VP
Joined: 09 Mar 2016
Posts: 1234

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19 Jul 2017, 10:40
Bunuel wrote:
rpm wrote:
Hello, a question in Manhattan GMAT guide.
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?
My ans: Total no. of ways to give gifts = 5P5 /2! = 5!/2! = 60.
How do I account for the youngest niece's condition?

Total # of ways to distribute SSEGT among 5 sisters (without restriction) is $$\frac{5!}{2!}=60$$;
The # of ways when the youngest niece gets G is: $$\frac{4!}{2!}=12$$ (give G to youngest and then distribute SSET among 4 sisters).

So, # of ways when youngest niece doesn't get G is: $$60-12=48$$.

Hope it's clear.
[/ hi Bunuel! Can you please explain why do you divide 5! by 2! what formula is it? if the Combination formula is N! / K!(N-K)! and permutation formula is N! / (N-K)! then which formula did you use? ] thank you for taking time to reply
VP
Joined: 09 Mar 2016
Posts: 1234

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19 Jul 2017, 10:47
Bunuel wrote:
verycoolguy33 wrote:
Hi Bunel ,
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. Bunuel - and if write like this 5! + 2! ? what would it mean in this case? thank you !!!!!!!!!!
Intern
Joined: 13 Jul 2018
Posts: 7

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24 Sep 2018, 00:03
I do not get why 5! should be divided by 2!, and not 5! divided by 2!3!.

In other problems I've studied, calculations are always n!/r!(n-r)!. Can anyone explain when to use these two types of calculations? Thank you.
Intern
Joined: 03 Oct 2018
Posts: 4

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08 Oct 2018, 08:50
It may ans just 48,, bcz 5!/2!-4!/2!=60-12= 48

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Re: Gordon buys 5 dolls for his 5 nieces. The gifts include two &nbs [#permalink] 08 Oct 2018, 08:50

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