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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 \(\frac{2}{15}\).
(2) The number of different 4-painting groups that could be selected is 210.
(1) The probability that two paintings from the collection, Sunrise Harbor and Winter Field, will both be selected is \(\frac{2}{15}\).
(2) The number of different 4-painting groups that could be selected is 210.
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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 \(\frac{2}{15}\).
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 \(C(n - 2, 2) = \frac{(n - 2)!}{2!(n - 4)!} = \frac{(n - 2)(n - 3)}{2}\).
The total number of 4-painting groups is \(C(n, 4) = \frac{n!}{4!(n - 4)!} = \frac{n(n - 1)(n - 2)(n - 3)}{24}\).
Therefore, we are given that:
\(\frac{\frac{(n - 2)(n - 3)}{2}}{\frac{n(n - 1)(n - 2)(n - 3)}{24}} = \frac{2}{15}\)
\(\frac{(n - 2)(n - 3)}{2} * \frac{24}{n(n - 1)(n - 2)(n - 3)} = \frac{2}{15}\)
\(\frac{6}{n(n - 1)} = \frac{1}{15}\)
\(90 = n(n - 1)\)
\(n^2 - n - 90 = 0\)
\((n - 10)(n + 9) = 0\)
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
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 \(\frac{2}{15}\).
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 \(C(n - 2, 2) = \frac{(n - 2)!}{2!(n - 4)!} = \frac{(n - 2)(n - 3)}{2}\).
The total number of 4-painting groups is \(C(n, 4) = \frac{n!}{4!(n - 4)!} = \frac{n(n - 1)(n - 2)(n - 3)}{24}\).
Therefore, we are given that:
\(\frac{\frac{(n - 2)(n - 3)}{2}}{\frac{n(n - 1)(n - 2)(n - 3)}{24}} = \frac{2}{15}\)
\(\frac{(n - 2)(n - 3)}{2} * \frac{24}{n(n - 1)(n - 2)(n - 3)} = \frac{2}{15}\)
\(\frac{6}{n(n - 1)} = \frac{1}{15}\)
\(90 = n(n - 1)\)
\(n^2 - n - 90 = 0\)
\((n - 10)(n + 9) = 0\)
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
