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# A nickel, a dime, and 2 quarters are arranged along a side

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Manager
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Joined: 03 Nov 2010
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GMAT Date: 10-15-2011
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Kudos [?]: 93 [0], given: 8

A nickel, a dime, and 2 quarters are arranged along a side [#permalink]

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03 Nov 2010, 05:21
Hi guys...
i had a problem and needed an insight on the same:

A nickel, a dime, and 2 quarters are arranged along a side of the table. if the quarter and the dime have to face heads up and nickel can face either heads up or tails up, how many different arrangements of coins are possible?

The answer says 4!/2!) * 2!

Can somebody please explain me how to arrive to this ans?

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Re: A nickel, a dime, and 2 quarters [#permalink]

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03 Nov 2010, 07:29
2
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Expert's post
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This post was
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krishnasty wrote:
Hi guys...
i had a problem and needed an insight on the same:

A nickel, a dime, and 2 quarters are arranged along a side of the table. if the quarter and the dime have to face heads up and nickel can face either heads up or tails up, how many different arrangements of coins are possible?

The answer says 4!/2!) * 2!

Can somebody please explain me how to arrive to this ans?

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THEORY.
Permutations of $$n$$ things of which $$P_1$$ are alike of one kind, $$P_2$$ are alike of second kind, $$P_3$$ are alike of third kind ... $$P_r$$ are alike of $$r_{th}$$ kind such that: $$P_1+P_2+P_3+..+P_r=n$$ is:

$$\frac{n!}{P_1!*P_2!*P_3!*...*P_r!}$$.

For example number of permutation of the letters of the word "gmatclub" is $$8!$$ as there are 8 DISTINCT letters in this word.

Number of permutation of the letters of the word "google" is $$\frac{6!}{2!2!}$$, as there are 6 letters out of which "g" and "o" are represented twice.

Number of permutation of 9 balls out of which 4 are red, 3 green and 2 blue, would be $$\frac{9!}{4!3!2!}$$.

So, # of arrangements of a nickel, a dime, and 2 quarters, or arrangements of 4 letters NDQQ out of which 2 Q's appear twice will be $$\frac{4!}{2!}$$. Next, as nickel can face either heads up or tails up then we should multiple this number by 2, so finally we'll get: $$\frac{4!}{2!}*2$$.

Hope it's clear.
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Manager
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Joined: 03 Nov 2010
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Kudos [?]: 93 [0], given: 8

Re: A nickel, a dime, and 2 quarters [#permalink]

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03 Nov 2010, 11:31
thanks a lot...
cleared the concept quite well!!
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Re: A nickel, a dime, and 2 quarters are arranged along a side [#permalink]

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13 Sep 2015, 10:56
shouldnt it be HTHH or HHHH, so 4!/3! +1 = 5
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Kudos [?]: 106061 [0], given: 11607

Re: A nickel, a dime, and 2 quarters are arranged along a side [#permalink]

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14 Sep 2015, 05:27
SahilKataria wrote:
shouldnt it be HTHH or HHHH, so 4!/3! +1 = 5

The number of arrangements of a nickel, a dime, and 2 quarters, or arrangements of 4 letters NDQQ out of which 2 Q's appear twice will be $$\frac{4!}{2!}$$. Next, as nickel can face either heads up or tails up then we should multiple this number by 2, so finally we'll get: $$\frac{4!}{2!}*2$$.
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Re: A nickel, a dime, and 2 quarters are arranged along a side [#permalink]

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14 Sep 2015, 11:19
1
KUDOS
i get ur point,
its like nickel head ort tale or a dime head or a tail, not just a head or a tail ...so definitely more than 5 cases.
Apreciate help, Thank u.
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Re: A nickel, a dime, and 2 quarters are arranged along a side [#permalink]

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17 Dec 2016, 08:24
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Re: A nickel, a dime, and 2 quarters are arranged along a side   [#permalink] 17 Dec 2016, 08:24
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