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The flag of country A has to contain 3 stripes of the same

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Senior Manager
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The flag of country A has to contain 3 stripes of the same [#permalink] New post 18 Feb 2005, 08:03
00:00
A
B
C
D
E

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(N/A)

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0% (00:00) correct 0% (00:00) wrong based on 0 sessions
The flag of country A has to contain 3 stripes of the same width, all of which have to be either vertical or horizontal. If the flag of country A must consist of the national colours, which are green, red, yellow, black, and blue, how many differen flags can be created?

(A) 24
(B) 48
(C) 60
(D) 120
(E) 240
Senior Manager
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 [#permalink] New post 18 Feb 2005, 08:21
My probability sucks but i think for this one it is
5!/(5-3)!=60
but this is different ways of choosing 3 colors, but since the question is how many different combinations are possible on the flag, i think it means that if i pick Black, Yellow, and Red (1 choice) then of this i can have 6 different types of flags right?
So i am confused if order matters or not.
Id still go for 60, it seems like there is more but i am a little lost
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 [#permalink] New post 18 Feb 2005, 09:12
see why E?
at most i thought if there are 6 different ways to arrange the colors on the flag (3!)
and it is 60 different ways of choosing the colors shouldnt it be more then??
what was your reasoning?
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 [#permalink] New post 18 Feb 2005, 14:07
Right times two for vertical and horizontal...got it thanks.
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 [#permalink] New post 18 Feb 2005, 19:38
hmmmm.....

tricky question. for a moment i thought its C. but flags can be different if order is different. hence there are 120 ways. D
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 [#permalink] New post 19 Feb 2005, 09:57
horizontal: 5 (for the first stripe 5 colors are possible) * 4 (for the second stripe 4 c are possible) * 3 (for the third stripe 3 c are possible) = 60
vertikal: 5*4*3 = 60

=> 120 D) !
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 [#permalink] New post 01 Mar 2005, 19:47
can someone suggest an easy, tried-and-tested way to distinguish between permutation and combination problems?
I mistakenly did 5c2 * 2 for this problem, which got me nowhere...
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 [#permalink] New post 01 Mar 2005, 20:44
But before going to end today I wanted to peep into this contagious forum :-)

Heres how I do....

5 colors and you need to pick 3 ----> 5 C 3 ways = 10

After picking 3 colors arranging them is 3! ways = 6

But we need to consider horizontal and vertical also = 2

So 10*6*2 =120


Thanks
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 [#permalink] New post 02 Mar 2005, 10:36
lastochka wrote:
can someone suggest an easy, tried-and-tested way to distinguish between permutation and combination problems?
I mistakenly did 5c2 * 2 for this problem, which got me nowhere...


Just ask yourself if you change the orders of two items, does it count as two outcomes for the question.
  [#permalink] 02 Mar 2005, 10:36
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