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

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...

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

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.

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.

...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.

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.

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

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.

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)

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