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Five tokensone red, one blue, one green, and two whiteare arranged

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Bunuel
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Five tokensone red, one blue, one green, and two whiteare arranged [#permalink] New post   04 Apr 2024, 01:30
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­Five tokens—one red, one blue, one green, and two white—are arranged in a row, one next to another. If the two white tokens are next to each other, how many arrangements according to color are possible?

(A) 12
(B) 16
(C) 20
(D) 24
(E) 30


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Re: Five tokensone red, one blue, one green, and two whiteare arranged [#permalink] New post   04 Apr 2024, 03:36
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­We have the following tokens to arrange:  W W G B R 

We are told that the two white tokens are next to one another in every arrangement. Given that the two tokens are identical and are always next to one another, we can group the two white tokens as a single token. We now view the tokens as W G B R. 

To arrange four different things: \(4! = 24\)

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Re: Five tokensone red, one blue, one green, and two whiteare arranged [#permalink] New post   09 Apr 2024, 19:23
Bunuel How do you know to do 4! instead of 5C3 (5!/3!2!) ?
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Re: Five tokensone red, one blue, one green, and two whiteare arranged [#permalink] New post   10 Apr 2024, 00:49
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iishim wrote:
­Five tokens—one red, one blue, one green, and two white—are arranged in a row, one next to another. If the two white tokens are next to each other, how many arrangements according to color are possible?

(A) 12
(B) 16
(C) 20
(D) 24
(E) 30

Bunuel How do you know to do 4! instead of 5C3 (5!/3!2!) ?

­Since the two white tokens need to be together, consider them as one unit {ww}. Thus, we'd have 4 units in total: {ww}, {r}, {b}, and {g}. The number of ways to arrange those is 4! = 24.

Answer: D.
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Re: Five tokensone red, one blue, one green, and two whiteare arranged [#permalink]
10 Apr 2024, 00:49
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