How many 4-digit numbers can be formed by using the digits 0

Question

How many 4-digit numbers can be formed by using the digits 0-9, so that the numbers contains exactly 3 distinct digits?

(A) 1944
(B) 3240
(C) 3850
(D) 3888
(E) 4216

tabsang · Accepted Answer

Well, here's my approach:

Case I:  
The repeated digit is the unit's digit.
So, the 1st, 2nd and 3rd digits can be selected in 9 x 9 x 8 ways, respectively.
Now the  4th digit  (unit's digit) can be either equal to the 1st, 2nd or 3rd digit.
Thus, in all we have:
9x9x8x 3

Case II:  
The repeated digit is the ten's digit.
So, the 1st, 2nd and 4th digits can be selected in 9 x 9 x 8 ways, respectively.
Now the  3rd digit  (ten's digit) can be either equal to the 1st or 2nd digit.
Thus, in all we have:
9x9x 2 x8

Case III:  
The repeated digit is the hundred's digit.
So, the 1st, 3rd and 4th digits can be selected in 9 x 9 x 8 ways, respectively.
Now the  2nd digit  (hundred's digit) is equal to the 1st digit.
Thus, in all we have:
9x 1 x9x8

In totality, we have 9x9x8(3+2+1) = 9x9x8x6 = 3888

Hope this helps.

mbaiseasy · Answer

How many ways to select 3 digits from 0-9?  =\frac{10!}{3!7!} = 120 
How many ways to select a repeating digits?  3 
How many ways to arrange {D1,D2,R,R}?  =\frac{4!}{2!}=12

=120*36 = 4320

Now we have 0-9 that could be the first digit. We cannot allow 0 to be the first digit. We know 0-9 will occur evenly as a first digit in 4320 counts.  =4320 - \frac{4320}{10} = 3888

Answer: D

tabsang · Answer

Well, I did dig deep into GMAT combinatorics and got really stuck into some questions. Out of the many solutions, here's one that uses your approach in principle (Sincere thanks to the expert who helped). Have a look: Solution: Out of the 4 digits, any 2 have to be the same. Number of ways this is possible: 4C2 = 6 . Consider one case: Tens digit and units digit are the same: Number of options for the thousands digit = 9. (Any digit 1-9) Number of options for the hundreds digit = 9. (Any digit 0-9 not yet chosen) Number of options for the tens digit = 8. (Any digit 0-9 not yet chosen) Number of options for the units digit = 1. (Must be the same as the tens digit) To combine the options above, we multiply: 9*9*8*1 = 648 . Other cases: #ways if the HUNDREDS digit and the UNITS digit are the same (9*9*8*1) #ways if the THOUSANDS digit and the UNITS digit are the same (9*9*8*1) #ways if the HUNDREDS digit and the TENS digit are the same (9*9*1*8) #ways if the THOUSANDS digit and the TENS digit are the same (9*9*1*8) #ways if the THOUSANDS digit and the HUNDREDS digit are the same (9*1*9*8) Total #ways = 648*6 = 3888. Sincerely hope this helps

If this brought a smile to your face, cleared the doubt clouds and made your day then a quick kudos and a big smilie is in place. Cheers, Taz P.S.: It feels great that I'm able to help & share in the same way that others have helped and shared with me. Cheers to gmatclub. Cheers to bb

HumptyDumpty · Answer

Could you describe how you have arrived at your answer?

HumptyDumpty · Answer

I think I agree with you.

Here’s my approach:

You have 3!=6 of possible placements of the doubled number, because you treat the four digit number with the doubling as a 3-different-digits-number without the doubling, simply ignoring the doubling (the glue method). So you can count the possible arrangements of X Y Z=3!=6, here (the 4-digits are only illustrative):

X Y Z Z or X Y Z
X Z Z Y or X Z Y
Z Z X Y or Y X Z
Z X Z Y or Y Z X
X Z Y Z or Z X Y
Z X Y Z or Z Y X

The number of possible numbers made up from digits 0-9 for each of the above possibilities is 9*9*8, i.e.:

X Y Z Z:
- for X – 9 digits from 1-9 as 0 would be indifferent in the first place,
- for Y – 9 digits from 0-9 except for thousands digit,
- for Z Z – 8 digits from 0-9 except for thousands digit and hundreads digit.

The same scheme applies to each of the 6 possible arrangements listed above, therefore:
9*9*8*6 = 3888. The answer is D.

I hope it's correct. As usual the hardest part was to bump on the idea, however schematic this problem was.

tabsang · Answer

I think there is a problem here with this approach.

Try doing the same for a 5-digit number with 4 distinct digits.
Using my approach, the answer is 9*9*8*7*(4+3+2+1) = 45360

If I use your approach, the answer is as follows:
4!=24 (using glue method)
And 9*9*8*7 choices for the four digits.
Answer in this case would be 9*9*8*7*24 = 108864

Seems to be some confusion here.

HumptyDumpty · Answer

You're right, my luck, my bad.
Pity, apparently I can't understand the case thorough at the moment.
I'd appreciate detailed troubleshooting to the approach, if someone loves combinatorics.
Kudo for you.

priyamne · Answer

Ans:

there are 3 cases:
1st case
The repeated digit is the unit's digit.
So, the 1st, 2nd and 3rd digits can be selected in 9 x 9 x 8 ways.
Now the unit's digit can be either equal to the 1st, 2nd or 3rd digit.
 we have:
9x9x8x3

2nd case:
The repeated digit is the ten's digit.
So, the 1st, 2nd and 4th digits can be selected in 9 x 9 x 8 ways, respectively.
Now the ten's digit can be either equal to the 1st or 2nd digit. we have:
9x9x2x8

3rd case:
The repeated digit is the hundred's digit.
So, the 1st, 3rd and 4th digits can be selected in 9 x 9 x 8 ways, respectively.
Now the hundred's digit is equal to the 1st digit.
 we have:
9x1x9x8

so total= 9x9x8(3+2+1) = 9x9x8x6 = 3888
answer is (D).

kunal555 · Answer

Fixing Thousand's place and starting with 1

1  1  _ _
The empty spaces can be filled in 9x8 = 72 ways.
and the hundred's, ten's and unit's place can be arranged in 3! way or 6 ways. So total combination for 1 will be 72*6 = 432

Similarly we can obtain combinations for all the remaining numbers. The thousand's place can take 9 values(since 0 cannot be at thousand's place). 
So total number of combinations will be
432*9 =3888

Answer:- D

LeonidK · Answer

thanks for explanations
missed case 2 and 3

Turchet · Answer

I am get a higher number than the answer, can someone please explain me my mistake?
I considered 2 ways: The first digit is being repeated   and 2 of the last three digits are equal  , so:
Lets say that B and C are numbers from 0-9, and A is a number between 1 and 9.
For    \rightarrow   
=  9 \cdot 1 \cdot 8 \cdot 8 \cdot 3! = 3888

For    / 
=  9 \cdot 9 \cdot 1 \cdot 8 \cdot \frac{3!}{2!} = 1944

Total = 3888 + 1944 = 5832.
I believe that I counting some combinations twice. But I cannot figure it out.
Thanks!

Ranan · Answer

Hi All, 
I am fairly new to this...but I solved it in a completely different approach...

First of all the 1st, 2nd and 3rd can be filled in 9x9x8=648 ways....then I just divided the options given in the question because after division, i will get an integer... 3888/648 = 6
so answer is D

took around 1:40 secs to solve it

Bismarck · Answer

OA:D

Case 1. Digits at thousands place repeated once 
X X _ _: thousands and hundreds digit same, other digits distinct 9*1*9*8
X _ X _: thousands and tens digit same, other digits distinct 9*9*1*8
X _ _ X: thousands and units digit same, other digits distinct 9*9*8*1
Total digits under case 1: 3*9*9*8

Case 2. Digits at hundreds place repeated once 
_ X X _: hundreds digit and tens digit same, other digits distinct 9*9*1*8
_ X _ X: hundreds digit and units digit same, other digits distinct 9*9*8*1
Total digits under case 2: 2*9*9*8

Case 3. Digits at tens place repeated once 
_ _ X X: tens digit and units digit same, other digits distinct 9*9*8*1

Total digits under case 3: 1*9*9*8

Total Number of digits possible: 3*9*9*8+2*9*9*8+1*9*9*8 =6*9*9*8 =3888

Prasannathawait · Answer

If other answer options are not divisible by 6 then yes you are correct.

onoviosu · Answer

I have spent a good thirty minutes reading your explanations but I cant seem to fully understand your solution. Why do you need three cases where the unit tens and hundred digit is the number that is repeated. Isn't having the 1st, 2nd and 3rd number multipled by a 3 counting the possibility that the repeated number is one of the first three already selected(Thousand or Hundreds or Tens)? Why is the second scenario in which case (9x9x2x8) needed. Wouldn't that be double counting in some form?

To me the answer should just be 
 the 1st, 2nd and 3rd digits can be selected in 9 x 9 x 8 ways, respectively. The last digit can be either one of the already selected options which is 3 possibilities. Option A (1944)

I hope my question is clear. Please advise.

ishant5243 · Answer

I am also not able to understand the exact logic. Can someone help me to understand why there are 2 possible ways to use an identical digit in ten's place and 1 way for hundredth's place.

Lanline · Answer

D. 3888 is the right answer.

We have to choose the 4 digit number, such that exactly 2 digits are same. Here are the 6 such combinations.

XXYZ, XYXZ, XYZX, YZXX,YXZX, YXXZ

Now, the numbers of way we can select X, Y and Z is 10, 9 and 8.

So, multiplying with the above 6 combinations the total such numbers are 10*9*8*6 = 4320.

But, if we take zero as the thousand digit, we have three digit number instead of 4. Such cases are 4320/10=432.

So, if we remove such numbers than the answer is 3888.

LaveenaPanchal · Answer

Hi,

Can I request an expert's reply.

Really finding it tough to understand that the answer is not 9 x 8 x 8 x 3 = 1944.

ZulfiquarA · Answer

I'm totally exhausted in finding a way how this question is to be done. If this question comes in GMAT I will take my whole Quant time.
Luckily I got the answer right when I first started doing it but the approach might be wrong as I have doubts. I'm sharing the approach others are telling as I find it more accurate, I will also share my first approach which I think might not be right.

Approach 1: Total elements - those which start with 0 and have 3 distinct elements
Find total numbers with distinct 3 elements and 1 common: 10*9*8*3/2=1080
the common element can be placed in 4 different ways as the 1st/2nd/3rd or 4th digit so Total elements= 1080*4=4320
Elements that have 0 as a distinct and common: 1*1*9*8=72, Now this can be rearranged in 6 ways = 72*6=432
So elements according to the condition minus those which start with zero: 4320-432=3888

Approach 2: Find the numbers of each scenario and add them
Total 4 digit numbers with distinct values: 9*9*8*3/2=972
3 can be rearranged in 4 ways: 972*4=3888
Even though I luckily got the answer right with this one I highly doubt it is right as when rearranging "3 element symbol" comes as first can be a 0 too.

I know you asked for an expert but until an expert answers it try to reason out with these approaches maybe you could find a way.

How many 4-digit numbers can be formed by using the digits 0

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How many 4-digit numbers can be formed by using the digits 0

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