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# There are 3 red chips and 2 blue ones. When arranged in a

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14 Feb 2010, 09:06
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bmwhype2 wrote:
There are 3 red chips and 2 blue ones.
When arranged in a row, they form a certain color pattern, for example RBRRB. How many color patterns?

10
12
24
60
100

$$\frac{5!}{3! * 2!} = 10$$ ... A
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08 Sep 2010, 02:53
bmwhype2 wrote:
There are 3 red chips and 2 blue ones.
When arranged in a row, they form a certain color pattern, for example RBRRB. How many color patterns?

10
12
24
60
100

This could be put in a format RRRBB. So total ways of choosing is 5! and since there are 3 R and 2 B, putting the R/B differently within each color pattern does not change anything so we need to remove them. So the formula is 5!/(3!2!) = 10

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Re: There are 3 red chips and 2 blue ones. When arranged in a [#permalink]

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24 Feb 2012, 12:07
Using MGMAT's anagram technique, ATLANTA is can be rearranged $$\frac{7!}{(3!)(2!)}$$ ways....coz of 3As and 2Ts.
Applying the same technique here:
RRRBB can be arranged in $$\frac{5!}{(3!)(2!)}$$ ways
=10
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Re: There are 3 red chips and 2 blue ones. When arranged in a [#permalink]

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06 Jun 2012, 10:25
5! for the permutation of number of ways it can be arranged. 2! and 3! for similar arrangements of Red and Black chips as no distinction factor between chips is mentioned i.e. all Red chips are identical and all black chips are identical too.

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Re: There are 3 red chips and 2 blue ones. When arranged in a [#permalink]

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09 Jul 2013, 14:53

$$C^5_2 * C^3_3 = 10$$

The 2 blue placed from the slot 1 to 5 and the remaining 3 placed in the rest 3 slots.
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Re: There are 3 red chips and 2 blue ones. When arranged in a [#permalink]

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09 Jul 2013, 15:12
Re: There are 3 red chips and 2 blue ones. When arranged in a   [#permalink] 09 Jul 2013, 15:12

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