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A nickel, a dime, and 2 identical quarters are arranged
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Author:  marcodonzelli [ 31 Jan 2008, 21:22 ]
Post subject:  A nickel, a dime, and 2 identical quarters are arranged

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

Author:  Bunuel [ 05 Feb 2012, 18:42 ]
Post subject:  Re: still permutations with repetitions

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.

Author:  T740qc [ 05 Feb 2012, 16:14 ]
Post subject:  Re: still permutations with repetitions

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!

Author:  Mbawarrior01 [ 13 Dec 2012, 08:09 ]
Post subject:  Re: still permutations with repetitions

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??

Author:  Bunuel [ 13 Dec 2012, 08:17 ]
Post subject:  Re: still permutations with repetitions

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.

Hope it's clear.

Author:  mbaiseasy [ 28 Dec 2012, 06:29 ]
Post subject:  Re: A nickel, a dime, and 2 identical quarters are arranged

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...

Answer: B

Author:  mvictor [ 06 Dec 2015, 10:51 ]
Post subject:  Re: A nickel, a dime, and 2 identical quarters are arranged

oh man...made a crucial mistake...
wrote 4!/2! * 2
and rewritten it as 4*3/1*2 and forgot to put on top 2!
:(
because of this, I got 12 :(

Author:  shalinkotia [ 07 May 2016, 00:42 ]
Post subject:  Re: A nickel, a dime, and 2 identical quarters are arranged

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).

Hence the answer = 4C2 * 2! * 4 = 48. Please suggest Bunuel.

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

Author:  Himanshu9818 [ 07 May 2016, 03:31 ]
Post subject:  Re: A nickel, a dime, and 2 identical quarters are arranged

[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

Answer B

Author:  chetan2u [ 07 May 2016, 03:39 ]
Post subject:  Re: A nickel, a dime, and 2 identical quarters are arranged

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).

Hence the answer = 4C2 * 2! * 4 = 48. Please suggest Bunuel.

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..

Author:  shalinkotia [ 12 May 2016, 09:08 ]
Post subject:  Re: A nickel, a dime, and 2 identical quarters are arranged

chetan2u - yes, misread the question. thanks.

Author:  geetgmat [ 11 Sep 2016, 02:57 ]
Post subject:  Re: A nickel, a dime, and 2 identical quarters are arranged

This clears it up!! Thanks a lot!!!

Posted from my mobile device

Author:  ashisplb [ 14 Jun 2017, 11:03 ]
Post subject:  Re: A nickel, a dime, and 2 identical quarters are arranged

Can anyone give a complete explanation for this

Sent from my ONE A2003 using GMAT Club Forum mobile app

Author:  Bunuel [ 14 Jun 2017, 11:06 ]
Post subject:  Re: A nickel, a dime, and 2 identical quarters are arranged

ashisplb wrote:
Can anyone give a complete explanation for this

Sent from my ONE A2003 using GMAT Club Forum mobile app


There are several solutions given above. You should be more specific when asking a question.

Author:  Leo8 [ 14 Jun 2017, 11:17 ]
Post subject:  Re: A nickel, a dime, and 2 identical quarters are arranged

+ B = 24

Attachment:
FullSizeRender (9).jpg
FullSizeRender (9).jpg [ 68.03 KiB | Viewed 17824 times ]

Author:  ashisplb [ 14 Jun 2017, 11:23 ]
Post subject:  Re: A nickel, a dime, and 2 identical quarters are arranged

Please let me know whether my approach is correct or not

For 2 identical quarters =4C2
For dime =2 heads up are available
Nickel =either heads up or tails up =2
Total=4C2*2*2=24

Sent from my ONE A2003 using GMAT Club Forum mobile app

Author:  Stonely [ 10 Dec 2019, 07:33 ]
Post subject:  Re: A nickel, a dime, and 2 identical quarters are arranged

easiest way to think about this is from the nickles position
nickles in position 1 as heads or tails and only the dime changes or N(2)QQD, N(2)QDQ, N(2)DQQ = 6
nickles in position 2 as heads or tails and only the dime changes or QN(2)QD, NQ(2)DQ, ND(2)QQ = 6
nickles in position 3 as heads or tails and only the dime changes or QQN(2)D, QDN(2)Q, DQN(2)Q = 6
nickles in position 4 as heads or tails and only the dime changes or QQDN(2), QDQN(2), DQQN(2) = 6
final count 24 seems slow but once you see the pattern it's easy

Author:  CEdward [ 29 Jan 2021, 13:51 ]
Post subject:  Re: A nickel, a dime, and 2 identical quarters are arranged

This is a combinations + permutations problem.

Combinations:
Nickle has two ways of facing - up or down
Quarters and dimes can only face one way.

2 x 1 x 1 x 1 = 2

Permutations: With repeats
4! / 2! = 4 x 3 = 12

Total number of arrangements = 2 x 12 = 24

Answer is B.

Author:  Fdambro294 [ 17 May 2021, 08:41 ]
Post subject:  Re: A nickel, a dime, and 2 identical quarters are arranged

(1st) Find the total no. of ways we can arrange the 4 coins into unique arrangements, given that the 2 quarters are identical element.

This would be the same no. of arrangement if we arranged the following word:

N - D - Q - Q

4! / (2! 1! 1!) = no. of unique arrangements of the 4 coins in the 4 spots

= 12 ways


AND

(2nd) For each one of these 12 ways, we can have the Nickel’s face showing Heads or Tails. Thus there are 2 ways to vary up the arrangement for the 12 ways

(12) * (2) =

24 ways

Posted from my mobile device

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