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A nickel, a dime, and 2 identical quarters are arranged [#permalink]
31 Jan 2008, 21:22

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A

B

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Difficulty:

65% (hard)

Question Stats:

49% (02:06) correct
51% (01:11) wrong based on 234 sessions

A nickel, a dime, and 2 identical quarters are arranged along a side of a table. If the quarters and the dime have to face heads up, while the nickel can face either heads up or tails up, how many different arrangements of coins are possible? A. 12 B. 24 C. 48 D. 72 E. 96

Re: still permutations with repetitions [#permalink]
05 Feb 2012, 16:14

can somone explain this please? 4!/2! to get to diff ways the coins can be placed taking into acct the repeat in quarters but i don't get the *2! part. Thanks!

Re: still permutations with repetitions [#permalink]
05 Feb 2012, 18:42

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Expert's post

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T740qc wrote:

can somone explain this please? 4!/2! to get to diff ways the coins can be placed taking into acct the repeat in quarters but i don't get the *2! part. Thanks!

Welcome to Gmat Club.

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 \(\frac{9!}{4!3!2!}\).

BACK TO THE ORIGINAL QUESTION: A nickel, a dime, and 2 identical quarters are arranged along a side of a table. If the quarters and the dime have to face heads up, while the nickel can face either heads up or tails up, how many different arrangements of coins are possible? A. 12 B. 24 C. 48 D. 72 E. 96

# 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: \(\frac{4!}{2!}*2=24\).

Re: still permutations with repetitions [#permalink]
13 Dec 2012, 08:09

Bunuel wrote:

T740qc wrote:

can somone explain this please? 4!/2! to get to diff ways the coins can be placed taking into acct the repeat in quarters but i don't get the *2! part. Thanks!

Welcome to Gmat Club.

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 \(\frac{9!}{4!3!2!}\).

BACK TO THE ORIGINAL QUESTION: A nickel, a dime, and 2 identical quarters are arranged along a side of a table. If the quarters and the dime have to face heads up, while the nickel can face either heads up or tails up, how many different arrangements of coins are possible? A. 12 B. 24 C. 48 D. 72 E. 96

# 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: \(\frac{4!}{2!}*2=24\).

Answer: B.

Hope it's clear.

I completely understood the concept. But i am little doubtful about one more condition given " If the quarters and dime have to face heads up" Do we assume that they are facing up or we should subtract the possibility of having tails up??

Re: still permutations with repetitions [#permalink]
13 Dec 2012, 08:17

1

This post received KUDOS

Expert's post

aditi2013 wrote:

Bunuel wrote:

T740qc wrote:

can somone explain this please? 4!/2! to get to diff ways the coins can be placed taking into acct the repeat in quarters but i don't get the *2! part. Thanks!

Welcome to Gmat Club.

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 \(\frac{9!}{4!3!2!}\).

BACK TO THE ORIGINAL QUESTION: A nickel, a dime, and 2 identical quarters are arranged along a side of a table. If the quarters and the dime have to face heads up, while the nickel can face either heads up or tails up, how many different arrangements of coins are possible? A. 12 B. 24 C. 48 D. 72 E. 96

# 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: \(\frac{4!}{2!}*2=24\).

Answer: B.

Hope it's clear.

I completely understood the concept. But i am little doubtful about one more condition given " If the quarters and dime have to face heads up" Do we assume that they are facing up or we should subtract the possibility of having tails up??

The nickel can face either heads up or tails up, thus we multiply the total # of ways in which we can arrange NDQQ by 2. The quarters and the dime have to face heads up, so only 1 choice for both of them, thus we don't need to multiply further.

Re: A nickel, a dime, and 2 identical quarters are arranged [#permalink]
28 Dec 2012, 06:29

marcodonzelli wrote:

A nickel, a dime, and 2 identical quarters are arranged along a side of a table. If the quarters and the dime have to face heads up, while the nickel can face either heads up or tails up, how many different arrangements of coins are possible? A. 12 B. 24 C. 48 D. 72 E. 96

How many ways to arrange {N}{D}{Q}{Q} ? \(=\frac{4!}{2!} = 12\)

\(=12*1*1*1*(2)\) Since there are two ways to arrange the nickel...

Re: A nickel, a dime, and 2 identical quarters are arranged [#permalink]
20 Jan 2014, 10:19

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Re: A nickel, a dime, and 2 identical quarters are arranged [#permalink]
24 Jan 2015, 20:31

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