Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

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

Show Tags

31 Jan 2008, 22:22

2

This post received KUDOS

6

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

65% (hard)

Question Stats:

51% (02:08) correct
49% (01:08) wrong based on 384 sessions

HideShow timer Statistics

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]

Show Tags

05 Feb 2012, 19:42

4

This post received KUDOS

Expert's post

5

This post was BOOKMARKED

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]

Show Tags

13 Dec 2012, 09:17

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]

Show Tags

28 Dec 2012, 07:29

1

This post received KUDOS

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]

Show Tags

07 May 2016, 04:39

1

This post received KUDOS

Expert's post

shalinkotia wrote:

Hi, shouldn't the answer be 48. For the Nickel and the Dime, we can select 2 places out of 4 in 4C2 ways followed by the arrangement in 2! ways. The two quarters then can have 4 combinations (HH, HT, TH, TT).

Please ignore any typos as I am new to this forum.

Hi, you have adopted the correct approach but misread the Q.. It says that quarters and dime have to be faced heads up.. ONLY Nickel can be either head or tail.. so Nickel can be placed in two ways.. ans 4C2*2!*2= 24.. _________________

Re: still permutations with repetitions [#permalink]

Show Tags

05 Feb 2012, 17:14

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]

Show Tags

13 Dec 2012, 09:09

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: A nickel, a dime, and 2 identical quarters are arranged [#permalink]

Show Tags

20 Jan 2014, 11:19

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________

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

Show Tags

24 Jan 2015, 21:31

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________

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

Show Tags

07 May 2016, 01:42

Hi, shouldn't the answer be 48. For the Nickel and the Dime, we can select 2 places out of 4 in 4C2 ways followed by the arrangement in 2! ways. The two quarters then can have 4 combinations (HH, HT, TH, TT).

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

Show Tags

07 May 2016, 04:31

[quote="marcodonzelli"]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

No of ways when nickel face head=4!/2!=12 no of ways when nickel face tail= 4!/2!=12 total no of ways =24

Part 2 of the GMAT: How I tackled the GMAT and improved a disappointing score Apologies for the month gap. I went on vacation and had to finish up a...

Cal Newport is a computer science professor at GeorgeTown University, author, blogger and is obsessed with productivity. He writes on this topic in his popular Study Hacks blog. I was...

So the last couple of weeks have seen a flurry of discussion in our MBA class Whatsapp group around Brexit, the referendum and currency exchange. Most of us believed...