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A nickel, a dime, and 2 identical quarters are arranged [#permalink]
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 31 Jan 2008, 22:22
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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
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Last edited by Bunuel on 05 Feb 2012, 19:44, edited 1 time in total.
Edited the question and added the OA
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Re: still permutations with repetitions [#permalink]
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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!
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Re: still permutations with repetitions [#permalink]
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05 Feb 2012, 19:42
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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: \(\frac{4!}{2!}*2=24\). Answer: B. Hope it's clear.
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Re: still permutations with repetitions [#permalink]
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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??
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Re: still permutations with repetitions [#permalink]
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13 Dec 2012, 09:17
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.
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Re: A nickel, a dime, and 2 identical quarters are arranged [#permalink]
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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
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Re: A nickel, a dime, and 2 identical quarters are arranged [#permalink]
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Re: A nickel, a dime, and 2 identical quarters are arranged [#permalink]
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24 Jan 2015, 21:31
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Re: A nickel, a dime, and 2 identical quarters are arranged [#permalink]
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06 Dec 2015, 11:51
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
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Re: A nickel, a dime, and 2 identical quarters are arranged [#permalink]
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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). Hence the answer = 4C2 * 2! * 4 = 48. Please suggest Bunuel. Please ignore any typos as I am new to this forum.
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Re: A nickel, a dime, and 2 identical quarters are arranged [#permalink]
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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
Answer B
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Re: A nickel, a dime, and 2 identical quarters are arranged [#permalink]
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07 May 2016, 04:39
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..
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Re: A nickel, a dime, and 2 identical quarters are arranged [#permalink]
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12 May 2016, 10:08
chetan2u - yes, misread the question. thanks.
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Re: A nickel, a dime, and 2 identical quarters are arranged [#permalink]
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11 Sep 2016, 03:57
This clears it up!! Thanks a lot!!!
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Re: A nickel, a dime, and 2 identical quarters are arranged [#permalink]
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14 Jun 2017, 12:03
Can anyone give a complete explanation for this Sent from my ONE A2003 using GMAT Club Forum mobile app
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Re: A nickel, a dime, and 2 identical quarters are arranged [#permalink]
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14 Jun 2017, 12:06
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Re: A nickel, a dime, and 2 identical quarters are arranged [#permalink]
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14 Jun 2017, 12:17
+ B = 24 Attachment: 
FullSizeRender (9).jpg [ 68.03 KiB | Viewed 180 times ]
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Re: A nickel, a dime, and 2 identical quarters are arranged [#permalink]
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14 Jun 2017, 12:23
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
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Re: A nickel, a dime, and 2 identical quarters are arranged
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14 Jun 2017, 12:23
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