p different prizes are hidden inside v vases. One of the vases can ho

Question

p different prizes are hidden inside v vases. One of the vases can hold up to two prizes, while the rest can hold only one prize. Are there more than 12 different possible arrangements of the prizes hidden in the vases? 
1) p = 3
2) v = 3

chetan2u · Answer

1) p=3...
If v is 2, number of arrangements...2 in one and third in second so choosing 2 out of 3 in 3C2 so 3C2=3<12
If v is say 10 ...
Single in each, choose 3 out of 10 vases in 10C3>12..
Add to this where 1 can hold two and the other can hold 1
Insufficient

2) v=3..
One can hold 2 others one each..
So max p =4..
2 in one and one each in other two...
Total number 4C2*2=12...
Multiplication by 2 is because the ones that can hold one each can interchange the prize..
Say 4 prizes A,B,C,D
So if 1 can hold 2 and 2 and 3 can hold one each..
1-AB,2-C,3-D
1-AB,2-D,3-C
1-AC,2-B,3-D
1-AC,2-D,3-B
And so on
So it can never be greater than 12
Ans is NO
Sufficient

B

pushpitkc · Answer

From the question stem. we have the following information:
1. There are p different prizes which need to be hidden inside v vases
2. The maximum number of prizes exactly one of the vases can hold is 2.
3. Every other vase(v - 1) holds one prize.

1. We are given that the number of prizes is 3.
There is a possibility that a particular vase does not hold any prize. The total
number of ways in which the prizes can be distributed in any number as the
number of vases is unknown. If we have 12 vases, there are definitely more
than 12 ways in which the prizes are arranged in the vases. However, if we
have 3 vases, there are a total of 6 ways of arranging the prizes.  (Insufficient)

2. If the number of vases is 3, the maximum number of prizes are 2 + 2*1 = 4.
For 4 prizes in 3 vases, the possibilities are 2 - 1 - 1. There can be 4c2 or 6 ways
of selecting prizes for the first vase and 2 ways in which we select prizes for the
second vase. The total ways of arranging the prices are 6*2 = 12. Therefore, we
never have more than 12 different possibilities  (Sufficient - Option B)

IanStewart · Answer

This question makes no sense, for a half dozen reasons. For one, it already tells you the prizes are hidden, so there is only one arrangement a priori. Then it's not clear if it's one specific vase (a large vase say) that can hold two prizes, or if any of the vases could be the vase that can hold two prizes. From the OA, I gather the former interpretation is correct, but that's not what I would think reading the question. And then it's not clear when we put two prizes in a vase if their order matters (is one prize on top of the other?). They use the word "arrangement", which has a technical meaning in math that suggests order does matter, but the OA suggests it does not. That's among other issues.

So when I read this question, I have no idea what it even means. No GMAT question will ever be worded like this - real GMAT combinatorics questions are always very carefully worded, and you will always know precisely what freedoms and restrictions you have.

prabsahi · Answer

Thanks for the insight.I got confused here.Ignoring this question.

harshilNori · Answer

Here we would need to assume there are a max of V + 1 prizes to get to the solution. Is that a valid assumption to make?

Crytiocanalyst · Answer

1) p = 3
Since the number of vases are not given the possibilities can exceed 12 combination or short of the 12 mark depending on the number of vases 
Clearly INsuff

2) v = 3
Now since the vase number is 3 we can safely assume the number of prices be 4 
the arrangemnt of prize in vases can be done in 4c2 ways=6 
and the remaining arrangement can be completed in 2 ways 
Making the total no of arrangements to be 12 
Therefore irrespective of the prize the total no of combination cannot exceed 12

Hence IMO B

Saupayan · Answer

Came here to say this. Makes 0 sense.

Rachitpaliwal10 · Answer

Wht can't we consider 1,1,1 and 2,1,1 both cases for 2nd statement?

p different prizes are hidden inside v vases. One of the vases can ho

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