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by ketkoag » Sat May 16, 2009 1:15 am
There are 5 cars to be displayed in 5 parking spaces with all the cars facing the same direction. Of the 5 cars, 3 are red, 1 is blue and 1 is yellow. If the cars are identical except for color, how many different display arrangements of the 5 cars are possible?
(A) 20
(B) 25
(C) 40
(D) 60
(E) 125

please explain as i selected the wrong answer..
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by aj5105 » Sat May 16, 2009 2:04 am
Imagine the scenario.

The first slot can be filled in 5 ways.
Second in 4 ways.
Third in 3 ways.
Fourth in 2 ways.
Fifth in 1 way.

That's 5 * 4 * 3 * 2 * 1 = 5! = 120 ways.

Now, we know there are three red cars. The 3 red cars are identical.It doesn't matter R1 is next to R2 or R2 is next to R3.So, we need to divide those possibilities from 120 ways (because 120 includes all possibilities).

120/3! = 120/6 = 20

(A)
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by ketkoag » Sat May 16, 2009 2:33 am
oh my god!! i did a silly mistake here.....
nyways, thanks for ur reply.....
:) :)
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by dadagreat » Sat May 16, 2009 3:06 am
this will be 5!/(3! * 1! * 1!)=20
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by dumb.doofus » Sat May 16, 2009 10:34 am
ketkoag wrote:oh my god!! i did a silly mistake here.....
nyways, thanks for ur reply.....
:) :)
Number of permutations of n-things, taken all at a time, in which ‘P’ are of one type, ‘q’ of them are of second-type, ‘r’ of them are of third-type, and rest are all different is given by :-
n!/(p! x q! x r! )

Remember the above and you can use it for such problems..
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