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20 Dec 2019, 01:19
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
35%
(medium)
Question Stats:
72% (01:53) correct
28%
(02:09)
wrong
based on 186
sessions
History
Date
Time
Result
Not Attempted Yet
Jean is arranging paint brushes used by artists for an impressionist's exhibit. He has 4 from Van Gogh, 5 from Monet, 3 from Manet, 4 from Degas, and 2 from Toulouse-Lautrec to choose from. All brushes from the same artist are considered identical for ordering. Jean wants to group all the brushes from Van Gogh together. How many ways can he arrange the paint brushes by the artist that used them?
(A) \(5!\)
(B) \(15!\)
(C) \(\frac{15!}{4!}\)
(D) \(\frac{11!}{5!3!4!2!}\)
(E) \(\frac{15!}{5!3!4!2!}\)
Are You Up For the Challenge: 700 Level Questions
(A) \(5!\)
(B) \(15!\)
(C) \(\frac{15!}{4!}\)
(D) \(\frac{11!}{5!3!4!2!}\)
(E) \(\frac{15!}{5!3!4!2!}\)
Are You Up For the Challenge: 700 Level Questions
20 Dec 2019, 02:10
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Bunuel
Jean is arranging paint brushes used by artists for an impressionist's exhibit. He has 4 from Van Gogh, 5 from Monet, 3 from Manet, 4 from Degas, and 2 from Toulouse-Lautrec to choose from. All brushes from the same artist are considered identical for ordering. Jean wants to group all the brushes from Van Gogh together. How many ways can he arrange the paint brushes by the artist that used them?
(A) \(5!\)
(B) \(15!\)
(C) \(\frac{15!}{4!}\)
(D) \(\frac{11!}{5!3!4!2!}\)
(E) \(\frac{15!}{5!3!4!2!}\)
Are You Up For the Challenge: 700 Level Questions
(A) \(5!\)
(B) \(15!\)
(C) \(\frac{15!}{4!}\)
(D) \(\frac{11!}{5!3!4!2!}\)
(E) \(\frac{15!}{5!3!4!2!}\)
Are You Up For the Challenge: 700 Level Questions
Explanation:
As all the brushes of Van Gogh brushes are together, we can count 1 brush
Van Gogh = 1
Monet = 5
Manet = 3
Degas = 4
Toulouse-Lautrec = 2
Total number of ways to arrange 15 brushes = (1 + 5 + 3 + 4 + 2)! = 15!
Since all brushes from the same artist are considered identical for ordering, we must take think into account.
Thus, the number of ways he can arrange the paint brushes by the artist = 15!/5!3!4!2!.
IMO-E
Please give kudos, If you find my explanation good enough 
globaldesi
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20 Dec 2019, 05:39
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Bunuel
Jean is arranging paint brushes used by artists for an impressionist's exhibit. He has 4 from Van Gogh, 5 from Monet, 3 from Manet, 4 from Degas, and 2 from Toulouse-Lautrec to choose from. All brushes from the same artist are considered identical for ordering. Jean wants to group all the brushes from Van Gogh together. How many ways can he arrange the paint brushes by the artist that used them?
(A) \(5!\)
(B) \(15!\)
(C) \(\frac{15!}{4!}\)
(D) \(\frac{11!}{5!3!4!2!}\)
(E) \(\frac{15!}{5!3!4!2!}\)
Are You Up For the Challenge: 700 Level Questions
(A) \(5!\)
(B) \(15!\)
(C) \(\frac{15!}{4!}\)
(D) \(\frac{11!}{5!3!4!2!}\)
(E) \(\frac{15!}{5!3!4!2!}\)
Are You Up For the Challenge: 700 Level Questions
consider all vangough paint brushes as tied in band since they always have to be together
so total brushes are 1(all van gugh as a bunch)+ 5+3+4+2 =15!
and since each brush by a painter is similar
hence divide by similar groups
\(\frac{15!}{!5!3!4!2!}\) =\(\frac{15!}{5!3!4!2!}\)
thus answer E
