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How many 4digit numbers can be formed by using the digits 0
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16 Dec 2012, 04:40
Question Stats:
24% (02:49) correct 76% (02:31) wrong based on 344 sessions
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How many 4digit numbers can be formed by using the digits 09, 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
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Re: How many 4digit numbers can be formed by using the digits 0
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16 Dec 2012, 08:47
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: 9x9x8x 3Case 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 2x8 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 1x9x8 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 4digit numbers can be formed by using the digits 0
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16 Dec 2012, 08:00
Could you describe how you have arrived at your answer?
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Re: How many 4digit numbers can be formed by using the digits 0
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16 Dec 2012, 13:57
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 3differentdigitsnumber 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 4digits 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 09 for each of the above possibilities is 9*9*8, i.e.: X Y Z Z:  for X – 9 digits from 19 as 0 would be indifferent in the first place,  for Y – 9 digits from 09 except for thousands digit,  for Z Z – 8 digits from 09 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 4digit numbers can be formed by using the digits 0
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17 Dec 2012, 01:39
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 3differentdigitsnumber 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 4digits 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 09 for each of the above possibilities is 9*9*8, i.e.:
X Y Z Z:  for X – 9 digits from 19 as 0 would be indifferent in the first place,  for Y – 9 digits from 09 except for thousands digit,  for Z Z – 8 digits from 09 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 5digit 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 4digit numbers can be formed by using the digits 0
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17 Dec 2012, 08:55
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 4digit numbers can be formed by using the digits 0
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17 Dec 2012, 09:24
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 19) Number of options for the hundreds digit = 9. (Any digit 09 not yet chosen) Number of options for the tens digit = 8. (Any digit 09 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 4digit numbers can be formed by using the digits 0
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18 Dec 2012, 00:36
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 4digit numbers can be formed by using the digits 0
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Updated on: 28 Dec 2012, 19:20
Thanks for this post! +1
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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.



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Re: How many 4digit numbers can be formed by using the digits 0
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28 Dec 2012, 19:18
tabsang wrote: How many 4digit numbers can be formed by using the digits 09, 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 09? \(=\frac{10!}{3!7!} = 120\) 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 09 that could be the first digit. We cannot allow 0 to be the first digit. We know 09 will occur evenly as a first digit in 4320 counts. \(=4320  \frac{4320}{10} = 3888\) Answer: D
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Re: How many 4digit numbers can be formed by using the digits 0
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24 Sep 2015, 10:50
tabsang wrote: How many 4digit numbers can be formed by using the digits 09, 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 Answer: D



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Re: How many 4digit numbers can be formed by using the digits 0
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05 Mar 2017, 01:58
thanks for explanations missed case 2 and 3



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Re: How many 4digit numbers can be formed by using the digits 0
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05 Jul 2017, 11:24
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 09, 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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Re: How many 4digit numbers can be formed by using the digits 0
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17 Aug 2018, 07:48
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



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How many 4digit numbers can be formed by using the digits 0
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17 Aug 2018, 08:33
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




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