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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?

Thanks in advance.. _________________

Appreciation in KUDOS please! Knewton Free Test 10/03 - 710 (49/37) Princeton Free Test 10/08 - 610 (44/31) Kaplan Test 1- 10/10 - 630 Veritas Prep- 10/11 - 630 (42/37) MGMAT 1 - 10/12 - 680 (45/34)

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?

So please: Provide answer choices for PS questions. Post new questions in a separate thread.

As for your question:

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: \(\frac{4!}{2!}*2\).

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

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14 Sep 2015, 05:27

Expert's post

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: \(\frac{4!}{2!}*2\). _________________

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

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14 Sep 2015, 11:19

1

This post received 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.

gmatclubot

Re: A nickel, a dime, and 2 quarters are arranged along a side
[#permalink]
14 Sep 2015, 11:19

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