A museum curator will select 4 paintings from a collection to feature

Question

A museum curator will select 4 paintings from a collection to feature in a special exhibition. How many paintings are in the collection?

(1) The probability that two paintings from the collection, Sunrise Harbor and Winter Field, will both be selected is 2/15.
(2) The number of different 4-painting groups that could be selected is 210.

Bunuel · Accepted Answer

GMAT Club Official Solution:

A museum curator will select 4 paintings from a collection to feature in a special exhibition. How many paintings are in the collection?

(1) The probability that two paintings from the collection, Sunrise Harbor and Winter Field, will both be selected is 2/15.

Let the total number of paintings be n.

Once those 2 paintings are definitely included, the other 2 paintings in the exhibition must be chosen from the remaining n - 2 paintings. So the number of favorable selections is C(n - 2, 2) = (n - 2)!/(2!(n - 4)!) = (n - 2)(n - 3)/2

The total number of 4-painting groups is C(n, 4) = n!/(4!(n - 4)!) = n(n - 1)(n - 2)(n - 3)/24

Therefore, we are given that:

((n - 2)(n - 3)/2)/(n(n - 1)(n - 2)(n - 3)/24)= 2/15

(n - 2)(n - 3)/2 *24/(n(n - 1)(n - 2)(n - 3))= 2/15

6/(n(n - 1))= 1/15

90 = n(n - 1)

n^2 - n - 90 = 0

(n - 10)(n + 9) = 0

So n = 10.

Sufficient.

(2) The number of different 4-painting groups that could be selected is 210.

This implies C(n, 4) = 210

There can be only one value of n, from which 210 groups of 4 could be selected. Sufficient.

Answer: D.

thisisakr · Answer

Given, 4 painting needs to be selected

Let the total number of paintings be n

So let’s check statement 2: it says total no of different 4 painting groups is 210

Basically saying nC4 = 210
n!/4! * n-4! =210
n * n-1 * n-2 * n-3 = 210 * 4!

Solving it we get n=10 ——-  sufficient

Statement 1: it says probability of selecting 4 painting when Sunrise and winter are both selected is 2/15

So we only need to select 2 other paintings from n-2 remaining painting as sunrise and winter are already selected

Total no of ways to select 4 painting from n painting is nC4
Total no of ways to select 2 paintings from n-2 painting is n-2 C 2

So P = n-2C2 / nC4 =2/15

SOLVING THIS we get n=10 ——-  sufficient

Answer D

Edskore · Answer

This one is testing Combinations and Probability together in a DS context. Both statements independently lead to n = 10, so the answer is D. Here's the full walkthrough.

Let n = number of paintings in the collection.

Statement 2: C(n, 4) = 210.
n!/(4! * (n-4)!) = 210. You can recognize C(10, 4) = 10*9*8*7/24 = 5040/24 = 210. So n = 10. Sufficient.

Statement 1: P(both Sunrise Harbor and Winter Field are selected) = 2/15.
Fix those two paintings as selected. We need to choose 2 more from the remaining (n - 2) paintings. Ways to do this: C(n-2, 2).
Total ways to pick any 4 from n: C(n, 4).
So the probability is C(n-2, 2) / C(n, 4) = 2/15.

Let's simplify:
C(n-2, 2) = (n-2)(n-3)/2
C(n, 4) = n(n-1)(n-2)(n-3)/24

So the ratio is   /   = 24 /   = 12 /  .

Setting that equal to 2/15:
12 /   = 2/15
n(n-1) = 90
n^2 - n - 90 = 0
(n - 10)(n + 9) = 0
n = 10 (since n must be a positive integer). Sufficient.

The thing I find genuinely satisfying about this problem is that the (n-2)(n-3) terms cancel cleanly in S1, leaving just n(n-1) to solve. If you don't simplify that ratio before setting it equal to 2/15, the algebra looks brutal but it isn't. Cancel first, then solve.

Answer: D.

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kshaurya · Answer

But how do I know that all the paintings are distinct? If there could be multiple copies of paintings then the answer shall be E, right?

Bunuel · Answer

I'd say that is a reasonable assumption here. In a museum collection, paintings chosen for a special exhibition are naturally understood as distinct works, so treating them as different is not a stretch.

A museum curator will select 4 paintings from a collection to feature

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A museum curator will select 4 paintings from a collection to feature

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