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Re: How many 4-digit numbers can be formed by using the digits 0 [#permalink]

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16 Dec 2012, 09:47

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HumptyDumpty wrote:

Could you describe how you have arrived at your 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: 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
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Re: How many 4-digit numbers can be formed by using the digits 0 [#permalink]

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16 Dec 2012, 14:57

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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.
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Re: How many 4-digit numbers can be formed by using the digits 0 [#permalink]

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17 Dec 2012, 02:39

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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

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.
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Re: How many 4-digit numbers can be formed by using the digits 0 [#permalink]

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17 Dec 2012, 09:55

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.
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Re: How many 4-digit numbers can be formed by using the digits 0 [#permalink]

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17 Dec 2012, 10:24

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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
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Re: How many 4-digit numbers can be formed by using the digits 0 [#permalink]

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18 Dec 2012, 01:36

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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).
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Re: How many 4-digit numbers can be formed by using the digits 0 [#permalink]

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28 Dec 2012, 20:18

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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. \(=4320 - \frac{4320}{10} = 3888\)

Re: How many 4-digit numbers can be formed by using the digits 0 [#permalink]

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18 Apr 2014, 17:49

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Re: How many 4-digit numbers can be formed by using the digits 0 [#permalink]

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24 Sep 2015, 11:09

Hello from the GMAT Club BumpBot!

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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

Re: How many 4-digit numbers can be formed by using the digits 0 [#permalink]

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10 Nov 2016, 09:02

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Re: How many 4-digit numbers can be formed by using the digits 0 [#permalink]

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05 Jul 2017, 12:24

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!

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