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# There are 4 quarters and 3 dimes placed at random in a line.

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Manager
Joined: 07 Jun 2006
Posts: 113
Followers: 2

Kudos [?]: 7 [0], given: 0

There are 4 quarters and 3 dimes placed at random in a line. [#permalink]  15 Oct 2006, 10:14
There are 4 quarters and 3 dimes placed at random in a line. The probability that the both extreme coins are dimes is

a) 1/4
b) 1/3
c) 1/7
d) 1/12
e) 1/2
Senior Manager
Joined: 08 Jun 2006
Posts: 342
Location: Washington DC
Followers: 1

Kudos [?]: 23 [0], given: 0

I think it is 1 / 7

Explanation â€“

4 quarters are identical and 3 dimes are identical
4 quarters and 3 dimes can be placed in (4 + 3) ! / ( 4! * 3 !) ways = 35 ways

Consider extreme 2 corners are 2 dimes. Then the remaining coins are 4 quarters and 1 dimes.
4 quarters and 1 dime can be arranged in (4 + 1) ! / ( 4! * 1 !) = 5

So the probability is = (Possible 5 cases) / (total cases 35) = 1/7
Manager
Joined: 01 Oct 2006
Posts: 242
Followers: 1

Kudos [?]: 5 [0], given: 0

Re: Quarters n Dimes [#permalink]  15 Oct 2006, 10:28
yeah, I also get 1/7.
Manager
Joined: 08 Jul 2006
Posts: 89
Followers: 1

Kudos [?]: 0 [0], given: 0

1/7 too

D _ _ _ _ _ D

3P2 * 5P5
------------
7P7
Senior Manager
Joined: 05 Oct 2006
Posts: 268
Followers: 1

Kudos [?]: 4 [0], given: 0

i think the qus is simply asking how to choose 2 quarters out of a total of 7 coins...and not about how other coins will be placed...

what u think??
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