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

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

Could you describe how you have arrived at your 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.

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.

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.

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).

Thanks for this post! +1

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

thanks for explanations
missed case 2 and 3

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 [1] and 2 of the last three digits are equal [2], so:
Lets say that B and C are numbers from 0-9, and A is a number between 1 and 9.
For [1] \(\rightarrow\) [ A A B C]
= \(9 \cdot 1 \cdot 8 \cdot 8 \cdot 3! = 3888\)

For [2] [A B B C]/[A B C C]
= \(9 \cdot 9 \cdot 1 \cdot 8 \cdot \frac{3!}{2!} = 1944\)

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

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

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

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

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.

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.

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.

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.