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Re M2216 [#permalink]
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16 Sep 2014, 01:16
Official Solution:How many threedigit integers greater than 710 are there such that all their digits are different? A. 198 B. 202 C. 207 D. 209 E. 212 First find how many integers between 700 and 999 are such that all their digits are different. We have : \(\text{(3 options for the first digit)}*\text{(9 options for the second digit)}*\text{(8 options for the third digit)} = 216\) numbers. Among these 216 numbers, 9 (701, 702, 703, 704, 705, 706, 708, 709, 710) are not bigger than 710. The answer to the question is therefore \(216  9 = 207\). Answer: C
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Re: M2216 [#permalink]
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28 Nov 2015, 05:53
i think you have missed one number 707
21610



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Re: M2216 [#permalink]
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28 Nov 2015, 07:27
ekia wrote: i think you have missed one number 707
21610 Hi, 216 is the total number where all digits are different.. 707 has 7 at two places, so 707 has not been taken as a part of these 216 numbers..
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Re: M2216 [#permalink]
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12 Mar 2016, 14:19
Hi guys, I understand the solution above but have trouble seeing what I did wrong  can you please help? If the first digit is 7: 2nd slot can be anything from #2 to #9 with the exception of 7, so that's 92+11=7; 3rd slot can be anything from #1 to #9 with the exception of #7 and the digit in the 2nd slot, so that's 91+12=7; so the total number of digit combinations is 7*7 = 49 numbers If the first digit is 8: 2nd slot can be anything from #0 to #9 with the exception of 9, so that's 90+11=9; 3rd slot can be anything from #0 to #9 with the exception of #8 and the digit in the 2nd slot, so that's 90+12=8; so the total number of digit combinations is 9*8 = 72 numbers If the first digit is 9: the combination is the same as if the first digit is 8. So, 72. Together there are 49 + 2*72 = 193 combinations ... What did I do wrong? Thank you!!
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Re M2216 [#permalink]
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28 May 2016, 06:35
I think this is a highquality question and I don't agree with the explanation. Question asks for threedigit integers such that all their digits are different, then below numbers should not be included right? example  777, 788, 799, 888, 899, 877 etc..



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Re: M2216 [#permalink]
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28 May 2016, 06:37
santanu1b wrote: I think this is a highquality question and I don't agree with the explanation. Question asks for threedigit integers such that all their digits are different, then below numbers should not be included right? example  777, 788, 799, 888, 899, 877 etc.. Yes, those numbers should not and are not included.
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Re: M2216 [#permalink]
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04 Dec 2016, 08:37
Bunuel,
I solved the question in the below way. Please correct me if wrong.
Per the question, we have 3 options(7,8,9) for the first digit.
1) 188 = 64  7 is the first digit here
2) 198 = 72  8 is the first digit here
3) 198 = 72  9 is the first digit here
So total is 64+72+72 = 208. However, in the first case we have to remove one number to account for 710. So answer is 207.



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Re: M2216 [#permalink]
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30 Nov 2017, 01:54
happyface101 wrote: Hi guys, I understand the solution above but have trouble seeing what I did wrong  can you please help?
If the first digit is 7: 2nd slot can be anything from #2 to #9 with the exception of 7, so that's 92+11=7; 3rd slot can be anything from #1 to #9 with the exception of #7 and the digit in the 2nd slot, so that's 91+12=7; so the total number of digit combinations is 7*7 = 49 numbers
If the first digit is 8: 2nd slot can be anything from #0 to #9 with the exception of 9, so that's 90+11=9; 3rd slot can be anything from #0 to #9 with the exception of #8 and the digit in the 2nd slot, so that's 90+12=8; so the total number of digit combinations is 9*8 = 72 numbers
If the first digit is 9: the combination is the same as if the first digit is 8. So, 72.
Together there are 49 + 2*72 = 193 combinations ...
What did I do wrong? Thank you!! I also approached the question in the above way. Bunuel, can you please explain?



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Re: M2216 [#permalink]
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04 Dec 2017, 15:45
jasanisanket24 wrote: happyface101 wrote: Hi guys, I understand the solution above but have trouble seeing what I did wrong  can you please help?
If the first digit is 7: 2nd slot can be anything from #2 to #9 with the exception of 7, so that's 92+11=7; 3rd slot can be anything from #1 to #9 with the exception of #7 and the digit in the 2nd slot, so that's 91+12=7; so the total number of digit combinations is 7*7 = 49 numbers
If the first digit is 8: 2nd slot can be anything from #0 to #9 with the exception of 9, so that's 90+11=9; 3rd slot can be anything from #0 to #9 with the exception of #8 and the digit in the 2nd slot, so that's 90+12=8; so the total number of digit combinations is 9*8 = 72 numbers
If the first digit is 9: the combination is the same as if the first digit is 8. So, 72.
Together there are 49 + 2*72 = 193 combinations ...
What did I do wrong? Thank you!! I also approached the question in the above way. Bunuel, can you please explain? Try to explain :  You did the correct calculation for 8 and 9 hundred digit.  Anyway, you miscalculated for the 7 hundred digit : don't forget the 0 number in the ten and unit digit. That's why, the number of digits possible is 1*8*7, not 1*7*7. After this, you must add manually the number from 710719 (which is 7 different number). Hope it helps.
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Re: M2216 [#permalink]
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30 Jan 2018, 23:02
Bunuel or All,
Why is 3*9*8 used to determine the amount of numbers that are from 700799 and with three different integers?



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Re: M2216 [#permalink]
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30 Jan 2018, 23:12



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Re: M2216 [#permalink]
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01 Feb 2018, 23:38
Understood. Thank you!



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Re: M2216 [#permalink]
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07 May 2018, 11:11
Great explanation in the original solution!!! There are usually many ways to solve a problem.
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