On her way home from a business trip, Janice remembers that she needs

Question

On her way home from a business trip, Janice remembers that she needs to buy each of her three children a souvenir. At the airport shop, the only items left are two hats, two t-shirts, two posters, and two keychains. How many combinations can she select from if she wants each child to receive a different item?

A. 24
B. 32
C. 40
D. 48
E. 56

A. 24
B. 32
C. 40
D. 48
E. 56

GMATinsight · Accepted Answer

to buy from   two hats, two t-shirts, two posters, and two keychains

Every child has to get different item so we need 3 out of 4 items first in 4C3 = 4 ways

Now from every item there are 2 ways to pick the item for one child

i.e. Total bways to pick items for children = 4C3 *2 *2 *2 = 4*8 = 32

Answer: Option B

Luckisnoexcuse · Answer

Hat T-shirt Poster Keychain

4C3 = 4
3! = 6
6*4=24

Alternatively
If A gets Hat
B gets T-shirt then last one has 4 options = 4
B has three options so total will be 12

Now A has 4 options
So the total becomes 12*4 = 48

Two answers...
Please help me understand what i am missing

SamBoyle · Answer

So you're assuming there is 2 different t shirts, 2 different key chains, 2 different posters and 2 different hats? I agree with your solution if that is the case. I assumed they were the same there there are 4 ways to select item 1, 3 ways to select item 2 and 2 ways to select item 3. Or 4X3X2 or 24.

I think Veritas should be clearer on if they mean 2 different items or not. There questions are getting a lot **** as I move through the practice tests. Why the hell not right 2 different hats, 2 different posters. I don't get how the whole question is writing poorly.

GMATinsight · Answer

Giving "Two different shirts" would have been awesome however the question always mentions if the objects are identical so in absence of it you should consider them different and it's not an assumption...

Samuelboyle96 · Answer

I guess that's just a GMAT quirk that I'll remember.

I read the items as being the same and found that to be the difficulty of the question. If you told me you had 2 baseballs, I'd assume they were the same baseball and wouldn't need you to tell me you had 2 identical baseballs. I really see no logic in the GMAT thinking it's necessary to say when things are identical but not to say when things are different. That's an esoteric logic system and I believe has only shown up on Veritas not the GMAT itself.

Posted from my mobile device

Kinshook · Answer

Option for 1st child = 8
Option for 2nd child = 6
Option for 3rd child = 4

Since there are 3! Ways to arrange them, Total options = 8*6*4/3! = 32

IMO B

Posted from my mobile device

cyryl5 · Answer

What I have done is..

All of the ways to select 3 from the 8 total number of gifts - all of the ways to give a child two of the same gifts.
Two of the kids can have the same shirt, poster, keychain, or hat. That leaves 1 child with 6 options and you must select one of them.

So

8 nCr 3 - 4*(6 nCr 1)

56 - 24 = 32

B

Balkrishna · Answer

Hi  GMATinsight ,

Don't we need to arrange selected 3 souvenirs in 3! because shuffling the souvenirs among children will create different combinations?

Thank you.

GMATinsight · Answer

Hi  Balkrishna

The problem is about Janice selecting different combinations for children and it is NOT about children getting the combinations in all possible ways.

So in current form, the problem is simply about her selecting 3 out of 4 souvenirs and does NOT expect Janice to distribute them among children.

If she were to distribute the souvenirs among children then certainly we must have multiplied 3!

I hope that clears your doubt.

ScottTargetTestPrep · Answer

She has 4C3 = 4 choices to choose 3 items out 4 unique items (hat, t-shirt, poster, and keychain). Since each unique item has 2 choices, she has 2 x 2 x 2 = 8 ways to choose any group of 3 unique items. Therefore, she has a total of 4 x 8 = 32 ways to choose 3 unique items for her 3 children.

Answer: B

Ankit_Raturi · Answer

Hi, could you please elaborate on how there are 3! ways?

Bunuel · Answer

The point is that 8*6*4 gives all combinations of a group of three gifts:

x, y, z
y, x, z
y, z, x
x, z, y
z, x, y
z, y, x

Since the above groups are the same, we need to divide 8*6*4 by 3! to get rid of the duplications.

On her way home from a business trip, Janice remembers that she needs

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