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Intern  Joined: 14 Jun 2011
Posts: 15
How many 4-digit numbers can be formed by using the digits 0  [#permalink]

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59 00:00

Difficulty:   95% (hard)

Question Stats: 25% (02:44) correct 75% (02:32) wrong based on 348 sessions

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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

I got (D) in a little over 3.5 minutes and I don't even know if it's right :O Intern  Joined: 14 Jun 2011
Posts: 15
Re: How many 4-digit numbers can be formed by using the digits 0  [#permalink]

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44
6
HumptyDumpty wrote:

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:
9x9x8x3

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:
9x9x2x8

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:
9x1x9x8

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

Hope this helps.

And P.S.: If you find this helpful please hit the kudos button. It'll be my first  ##### General Discussion
Manager  Joined: 12 Dec 2012
Posts: 148
Location: Poland
Re: How many 4-digit numbers can be formed by using the digits 0  [#permalink]

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Manager  Joined: 12 Dec 2012
Posts: 148
Location: Poland
Re: How many 4-digit numbers can be formed by using the digits 0  [#permalink]

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6
1
2
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.
Intern  Joined: 14 Jun 2011
Posts: 15
Re: How many 4-digit numbers can be formed by using the digits 0  [#permalink]

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3
HumptyDumpty wrote:
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

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.
Manager  Joined: 12 Dec 2012
Posts: 148
Location: Poland
Re: How many 4-digit numbers can be formed by using the digits 0  [#permalink]

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1
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.
Intern  Joined: 14 Jun 2011
Posts: 15
Re: How many 4-digit numbers can be formed by using the digits 0  [#permalink]

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12
7
HumptyDumpty wrote:
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.

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
Intern  Joined: 24 Apr 2012
Posts: 44
Re: How many 4-digit numbers can be formed by using the digits 0  [#permalink]

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2
1
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
Senior Manager  Joined: 13 Aug 2012
Posts: 386
Concentration: Marketing, Finance
GPA: 3.23
Re: How many 4-digit numbers can be formed by using the digits 0  [#permalink]

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1
Thanks for this post! +1

Originally posted by mbaiseasy on 27 Dec 2012, 04:18.
Last edited by mbaiseasy on 28 Dec 2012, 19:20, edited 1 time in total.
Senior Manager  Joined: 13 Aug 2012
Posts: 386
Concentration: Marketing, Finance
GPA: 3.23
Re: How many 4-digit numbers can be formed by using the digits 0  [#permalink]

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13
3
tabsang wrote:
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

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. $$=4320 - \frac{4320}{10} = 3888$$

Manager  Joined: 29 Jul 2015
Posts: 150
Re: How many 4-digit numbers can be formed by using the digits 0  [#permalink]

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2
tabsang wrote:
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

I got (D) in a little over 3.5 minutes and I don't even know if it's right :O 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

Manager  G
Joined: 28 Jul 2016
Posts: 128
Re: How many 4-digit numbers can be formed by using the digits 0  [#permalink]

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thanks for explanations
missed case 2 and 3
Intern  B
Joined: 13 May 2017
Posts: 8
Re: How many 4-digit numbers can be formed by using the digits 0  [#permalink]

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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  $$\rightarrow$$ [ A A B C]
= $$9 \cdot 1 \cdot 8 \cdot 8 \cdot 3! = 3888$$

For  [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!
Intern  B
Joined: 27 Jun 2018
Posts: 2
Re: How many 4-digit numbers can be formed by using the digits 0  [#permalink]

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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

took around 1:40 secs to solve it
Manager  D
Joined: 18 Jun 2018
Posts: 247
How many 4-digit numbers can be formed by using the digits 0  [#permalink]

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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
Senior Manager  G
Joined: 10 Aug 2018
Posts: 280
Location: India
Concentration: Strategy, Operations
WE: Operations (Energy and Utilities)
Re: How many 4-digit numbers can be formed by using the digits 0  [#permalink]

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

Ranan wrote:
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

took around 1:40 secs to solve it

_________________
On the way to get into the B-school and I will not leave it until I win. WHATEVER IT TAKES.

" I CAN AND I WILL"
Intern  B
Joined: 19 Mar 2020
Posts: 6
Re: How many 4-digit numbers can be formed by using the digits 0  [#permalink]

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tabsang wrote:
HumptyDumpty wrote:

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:
9x9x8x3

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:
9x9x2x8

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:
9x1x9x8

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

Hope this helps.

And P.S.: If you find this helpful please hit the kudos button. It'll be my first  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) Re: How many 4-digit numbers can be formed by using the digits 0   [#permalink] 10 May 2020, 17:38

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