A nickel, a dime, and 2 identical quarters are arranged

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

Show Answer

Official Answer

B

A. 12
B. 24
C. 48
D. 72
E. 96

Bunuel · Accepted Answer

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.

T740qc · Answer

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!

Mbawarrior01 · Answer

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??

Bunuel · Answer

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.

Hope it's clear.

mbaiseasy · Answer

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

mvictor · Answer

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

shalinkotia · Answer

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.

chetan2u · Answer

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..

Leo8 · Answer

+ B = 24 FullSizeRender (9).jpg

Stonely · Answer

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

GulfTube · Answer

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

Show Answer

Official Answer

B

Hq Hq Hd Hn = 1C1*1C1*1C1*1C1 * 4!/2! = 12
or Hq Hq Hd Tn = 1C1*1C1*1C1*1C1 * 4!/2! = 12
12+12 = 24 
 B

kaustubhkohli12 · Answer

Hi , Just one question, if it had been either heads on all coin. we both have taken half of 12 in that case. So i am struggling to understand why not the same approach . In half cases it will be head and half it will be tail

Bunuel · Answer

The 4!/2! = 12 arrangements count only the  order of the coins , not whether they show heads or tails. Therefore, requiring all coins to be heads does  not  reduce 12 by half, since tails were never included in those 12 arrangements.

architkap · Answer

why are we treating this as a repetition of letters problem? Regardless of repetition of quarters they still need to be arranged right? Why is it not 4! * 2? (4 coins arranged in any order out of which nickles have 2 ways of being placed)

Bunuel · Answer

Because the two quarters are identical, exchanging them does not create a new arrangement. Therefore, the number of distinct arrangements is 4!/2!, which is then multiplied by 2 for the nickel’s two possible faces.

Laza · Answer

I like to solve it using the slots method, in which you do not even need the permutation formula.

There are 4 slots to the coins to occupy:

_ _ _ _ (4 slots)

The nickel has 4 possibilities; then, the dime has 3 remaining; after that, the 2 quarters have 2 slots to occupy, but the combination will be the same (because they are identical), so 1 option.

4x3x1 = 12

The nickel can be flipped, so that's 12*2 = 24

Hope it helps!

A nickel, a dime, and 2 identical quarters are arranged

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Problem Solving (PS) Questions

A nickel, a dime, and 2 identical quarters are arranged

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards