AAPL wrote:Economist GMAT

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. \(\frac{15!}{5!3!4!2!}\)

OA E

Since all the brushes of Van Gogh brushes are together, we can count 1 brush

So, we have

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!

However, 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 = \(\frac{15!}{5!3!4!2!}\).

The correct answer: E

Hope this helps!

-Jay
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AAPL wrote:Economist GMAT

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. \(\frac{15!}{5!3!4!2!}\)

OA E

[/quote]

The brushes can be arranged as follows:

[V-V-V-V]-MO-MO-MO-MO-MO-MA-MA-MA-D-D-D-D-T-T

Since the Van Gogh's must be together, we consider those as "one" brush, so we are actually ordering 15 items, not 18. Additionally, since the brushes from the other artists are indistinguishable, we divide by the number of those indistinguishable brushes. Thus, the total number of arrangements becomes

15!/(5!3!4!2!)

Answer: E