How many distinct four-digit numbers can be formed by the

Question

How many distinct four-digit numbers can be formed by the digits {1, 2, 3, 4, 5, 5, 6, 6}?
A. 280
B. 360
C. 486
D. 560
E. 606

For more on difficult counting problems, as well as the OE of this problem, see:
https://magoosh.com/gmat/2013/difficult- ... -problems/

Mike :-)

PiyushK · Accepted Answer

Quite easy:
X X X X 
Case 1: all different digits from 1,2,3,4,5,6 = 6x5x4x3= 360
Case 2: Double 5 and double 6 in any sequence 5566 or 5656 etc.. 4!/(2!x2!) = 6 
Case 3 : Double 5 or double 6 and other two digits distinct 55 _ _ or 66 _ _ (select two digit from remaining 5 digits) 
5c2 = 10 + 5c2=10 = 20 for both 55_ _ and 66 _ _
20 is just selection not sequences thus we should multiply it with 4!/2! = 12 -- 4! divide by 2! because 55 or 66 are duplicate digits.

Total sequences in case 3 = 20 x 12 = 240

Total= 240+6+360 = 606 (E)

mikemcgarry · Answer

Dear Dave, My friend, with all due respect, you did nine scenarios, but one can just do it in four, two of which are identical Case One --- four different digits (6C4 possibilities) Case Two --- two 5's and two distinct digits Case Three --- two 6's and two distinct digits (has to be equal to case #2; no need to re-calculate) Case Four --- two 5's and two 6's (4C2 possibilities) In other words, it can be done with conisderably less work. This is a problem that can be done with a mountain of work or, with an elegant solution, not that much. I believe the GMAT is rather fond of such questions. Mike :-)

DDB1981 · Answer

not sure if this is fair game on the GMAT but here's my answer:

the trick is isolating the repeating numbers (5 and 6), because these are the toughest to deal with. Based on this, there are 9 scenarios, but before we dive into them, please keep in mind a few things:

the word AND in counting problems is "multiplication" in math
when the word choose appears below, you're using the CHOOSE formula
when the word permute appears below, you're using the PERMUTATIONS formula

without further ado, the 9 scenarios are:

5 appears 0 times, 6 appears 0 times   --> choose 4 from the original 1,2,3,4 --> 4! = 24 ways to do this
5 appears 0 times, 6 appears 1 times   --> choose a spot for the "6", AND permute 3 from the original 1,2,3,4 --> 4 spots for "6" AND 24 permutations of 1,2,3,4 = 4*24 = 96
5 appears 0 times, 6 appears 2 times   --> choose 2 spots for the "6" AND permute 2 from the original 1,2,3,4 --> 6 spots for "6" AND 12 permutations of 1,2,3,4 = 6*12 = 72

5 appears 1 times, 6 appears 0 times   --> same math as   --> 96
5 appears 1 times, 6 appears 1 times   --> choose spots for the "6" AND "5", AND permute 2 from the original 1,2,3,4 --> 12 spots for "6" and "5" AND 12 permutations of 1,2,3,4 = 12*12 = 144
5 appears 1 times, 6 appears 2 times   --> choose 1 spot for "5" and 2 for "6" AND permute 1 from the original 1,2,3,4 --> 12 spots for "5" and "6" AND 4 permutations of 1,2,3,4 = 12*4=48

5 appears 2 times, 6 appears 0 times   --> same math as   --> 72
5 appears 2 times, 6 appears 1 times   --> same math as   --> 48
5 appears 3 times, 6 appears 2 times   --> choose 2 spots for the "6" and 2 for the "5"'s --> 6

now add those numbers together: 24+96+72 + 96+144+48 + 72+48+6 = 606

again, this is A LOT of work and not required, IMHO, in the GMAT, i'm going to go lie down now.
Dave

DDB1981 · Answer

Dear Mike,

thanks for posting back, I confess I don't really understand your explanation:

Firstly, your cases 2 and 3 seem to imply that you're choosing 2 from 6, then choosing another 4, implying that you're choosing 6 digits total

Secondly, while the math on scenario 4 is correct, I don't understand how you liken choosing two 5's and two 6's to choosing two items of four, the math here would be 4!/(2!2!), or, 4P4/(2P2*2P2) in your terminology

Finally, and perhaps, most importantly, the math doesn't add up to 606

Please advise, thanks,
Dave

PS: it's been my experience that, while choosing and redundancies both appear on the GMAT, they don't appear together. That is, questions requiring one to choose a subset of items from a set don't feature redundancies, and questions with redundant items don't include choosing a subset from a set. Great question though, very tough.

hungta · Answer

(1) How many different groups of 4 digits we can have from original group?

- Option 1: how many different groups of 4 different digits? -> consider a subgroup (1,2,3,4,5,6): we have (6C4)=15 groups, no group has two 5's or two 6's
- Option 2: how many different groups of 4 digits which contain two 5's -> consider a subgroup (1,2,3,4,6): we have (5C2)=10 groups
- Option 3: how many different groups of 4 digits which contain two 6's -> consider a subgroup (1,2,3,4,5): we have (5C2)=10 groups
- Option 4: how many differrent groups of 4 digits which contain two 5's and two 6's -> obviously, only 1
In total: we have 36 different groups of 4 digits.

(2) How many four-digit numbers we can have from 36 groups of 4 digits?
- Option 1: each group of 4 different digits will produce 4!=24 four-digit numbers
- Option 2: each group of 4 digits that contains two 5's will produce 4!/2=12 four-digit numbers
- Option 3: same with option 2 -> 4!/2=12 four-digit numbers
- Option 4: 4!/4=6 four-digit numbers

In total, there are: 15*24 + 10*12 + 10*12 + 1*6 = 606 four-digit numbers.

Hope it helps.

Hung

mikemcgarry · Answer

Dave, Since I have already spelled out a complete text explanation of my calculation in the blog article to which I linked in the head post, I am going to suggest that you go to that blog, where the same question is posted, and read the complete TE there toward the bottom of the article. You may even find the article helpful. Please let me know if you have any further questions. Mike :-)

kanigmat011 · Answer

How many distinct four-digit numbers can be formed by the digits {1, 2, 3, 4, 5, 5, 6, 6}?

A. 280

B. 360

C. 486

D. 560

E. 606

Bunuel · Answer

Merging topics.

Please search before posting and provide OAs with questions. Thank you.

thefibonacci · Answer

All different --> C(6,4)*4! = 360 
2 same and 2 different --> C(2,1)*C(5,2)*4!/2! = 240 
2 same and 2 same --> 4!/2!^2 = 6

total = 606

hatemnag · Answer

Hi Bunuel
Actually i did not see clear explanation for the OA. 
please help to start from.

mikemcgarry · Answer

Dear hatemnag , I'm happy to respond. :-)

I'm going answer in the place of the genius Bunuel because I am the author of this particular question. It is question #2 on this blog: https://magoosh.com/gmat/2013/difficult- ... -problems/ You will find the full OE there. Let me know if you have any more questions. Mike :-)

hkkat · Answer

(1,2,3,4,5,5,6,6)
there are 2 5s and 2 6s
  6P4 =360
  2C2 * 4!/2! 2! = 6
  2C1 *5C2 *4!/2! = 240

total: 360 + 6 + 240 = 606//
answer is E

tanghoathien · Answer

Hi  mikemcgarry

How many distinct four-digit numbers can be formed? --> do we need all 4 digits are distinct, thanks!

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