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# M22-16

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Math Expert
Joined: 02 Sep 2009
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16 Sep 2014, 01:16
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Difficulty:

95% (hard)

Question Stats:

49% (01:20) correct 51% (02:28) wrong based on 105 sessions

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How many three-digit integers greater than 710 are there such that all their digits are different?

A. 198
B. 202
C. 207
D. 209
E. 212
[Reveal] Spoiler: OA

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16 Sep 2014, 01:16
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Official Solution:

How many three-digit 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 $$216 - 9 = 207$$.

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28 Nov 2015, 05:53
i think you have missed one number 707

216-10
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Joined: 02 Aug 2009
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28 Nov 2015, 07:27
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Expert's post
ekia wrote:
i think you have missed one number 707

216-10

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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Absolute modulus :http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html

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12 Mar 2016, 14:19
1
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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 9-2+1-1=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 9-1+1-2=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 9-0+1-1=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 9-0+1-2=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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28 May 2016, 06:35
I think this is a high-quality question and I don't agree with the explanation. Question asks for three-digit 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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28 May 2016, 06:37
santanu1b wrote:
I think this is a high-quality question and I don't agree with the explanation. Question asks for three-digit 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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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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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 9-2+1-1=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 9-1+1-2=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 9-0+1-1=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 9-0+1-2=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.

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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 9-2+1-1=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 9-1+1-2=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 9-0+1-1=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 9-0+1-2=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.

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 710-719 (which is 7 different number).

Hope it helps.
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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 700-799 and with three different integers?
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Joined: 02 Sep 2009
Posts: 44298

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30 Jan 2018, 23:12
mrosale2 wrote:
Bunuel or All,

Why is 3*9*8 used to determine the amount of numbers that are from 700-799 and with three different integers?

(3 options for the first digit): 7, 8, or 9.

(9 options for the second digit): all digits except the one digit we used above.

(8 options for the third digit): all digits except the two digits we used above.

Check Constructing Numbers, Codes and Passwords Questions in our Special Questions Directory.

Hope it helps.
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Joined: 06 Jun 2016
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01 Feb 2018, 23:38
Understood. Thank you!
Re: M22-16   [#permalink] 01 Feb 2018, 23:38
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# M22-16

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