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# Q. There are 5 parking spots and 5 cars (3 blue, 1 red, and

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Q. There are 5 parking spots and 5 cars (3 blue, 1 red, and [#permalink]  07 Jul 2007, 10:56
Q. There are 5 parking spots and 5 cars (3 blue, 1 red, and 1 green). In how many ways could the cars be parked?

A. 120 B. 33 C. 20 D. 15 E. 6

Answer-C. First you can find the number of combinations for 5 cars if they all were of different colors and that will be 5! = 120. Then to apply the limiting condition, we need to divide the above number by the times that the 3 cars of the same color repeat, which is 3! So, the final answer we will get is 20

... Can somebody please explain this?

Thanks!
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I will assume that you understand the 5! = 120 part. If all the cars were in diffrent colors then this should have been the outcome (i.e. BGRYO)

but since you have BBBRG (repeting B color three times) and since B2B1 is the same as B1B2, you need to narrow your 5! by removeing the repeating outcomes.

The best way is to divide by the meaningless outcomes (like B1B2 = B2B1)

let's see a simple example:

two colors for three cars:

B = blue
G1,G2 = green

3! = 6

BG1G2
G1BG2
G1G2B
BG2G1
G2BG1
G2G1B

n = 6

-----------------------------------------------------------------------------

but once we decide that G1=G2

the outcome is 3!/2! = 3 (total / repeated)

BGG
GGB
GBG

n = 3

5!/3! (total / repeated)

n = 120/6 = 20

Last edited by KillerSquirrel on 07 Jul 2007, 13:55, edited 1 time in total.
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Thanks, that makes sense now. The fact that B1=B2=B3 wasn't that apparent in the question stem. Thanks anyway, much appreciated.
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