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How many 3 digit integers can be chosen such that none of
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Updated on: 27 Jan 2012, 08:35
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How many 3 digit integers can be chosen such that none of the digits appear more than twice and none of the digits equal zero ? A. 729 B. 720 C. 648 D. 640 E. 576 what is the wrong of my method which is : 3*(9*9*8) where first digit can be chosen by 9 ways second digit can be chosen by 9 ways third digit can be chosen by 8 ways and since it's anagram of ( SSD ) it can be arranged in 3 ways so the answer IMO is 3*(9*9*8)= 1944 which is wrong
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Originally posted by Silver89 on 27 Jan 2012, 07:42.
Last edited by Silver89 on 27 Jan 2012, 08:35, edited 1 time in total.



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Re: How many 3 digit integers can be chosen such that none of
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27 Jan 2012, 08:40
Thanks, In fact you can solve it in another way too which is number of ways without restriction = 9*9*9 = 729 number of ways that break restriction = 9*1*9 = 9 so the answer is 7299 = 720 My question why can't i say first digit can be chosen by 9 ways second digit can be chosen by 9 ways third digit can be chosen by 8 ways and we can arrange it in 3 ways, thus the answer is 9*9*8*3
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How many 3 digit integers can be chosen such that none of
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27 Jan 2012, 09:24
Silver89 wrote: How many 3 digit integers can be chosen such that none of the digits appear more than twice and none of the digits equal zero ?
A. 729 B. 720 C. 648 D. 640 E. 576
Thanks, In fact you can solve it in another way too which is number of ways without restriction = 9*9*9 = 729 number of ways that break restriction = 9*1*9 = 9 so the answer is 7299 = 720
My question why can't i say first digit can be chosen by 9 ways second digit can be chosen by 9 ways third digit can be chosen by 8 ways and we can arrange it in 3 ways, thus the answer is 9*9*8*3 Ah, I got your question. Yes, the easiest way would be as you pointed out: A. Numbers without restriction = 9^3; B. 3digit integers with all alike digits = 9 (111, 222, ..., 999); AB=9^39=720. Now, if we go your way it should be 9*8*7+9*8*3=720, where 9*8*7 is # of 3digit integers with all distinct digits and 9*8*3 is # of 3digit integers with two alike digits (XXY  9*8 and multiplied by 3!/2! to get different permutations). So, as there are two cases you cannot combine them in one formula and apply one factorial correction (3!/2!) to it. Hope it's clear.
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Re: How many 3 digit integers can be chosen such that none of
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29 Nov 2014, 01:40
Bunuel wrote: Silver89 wrote: Thanks, In fact you can solve it in another way too which is number of ways without restriction = 9*9*9 = 729 number of ways that break restriction = 9*1*9 = 9 so the answer is 7299 = 720
My question why can't i say first digit can be chosen by 9 ways second digit can be chosen by 9 ways third digit can be chosen by 8 ways and we can arrange it in 3 ways, thus the answer is 9*9*8*3 Ah, I got your question. Yes, the easiest way would be as you pointed out: A. Numbers without restriction = 9^3; B. 3digit integers with all alike digits = 9 (111, 222, ..., 999); AB=9^39=720. Now, if we go your way it should be 9*8*7+9*8*3=720, where 9*8*7 is # of 3digit integers with all distinct digits and 9*8*3 is # of 3digit integers with two alike digits (XXY  9*8 and multiplied by 3!/2! to get different permutations). So, as there are two cases you cannot combine them in one formula and apply one factorial correction (3!/2!) to it. Hope it's clear. Sorry I don't understand both ways you mentioned. I dont know what is wrong with my way 9 ways for the first digit 9 ways for the second digit 8 ways for the last one => 9 x 9 x 8 = 648 ways. please explain! Many thanks



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Re: How many 3 digit integers can be chosen such that none of
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29 Nov 2014, 05:39
HuongShashi wrote: Bunuel wrote: Silver89 wrote: Thanks, In fact you can solve it in another way too which is number of ways without restriction = 9*9*9 = 729 number of ways that break restriction = 9*1*9 = 9 so the answer is 7299 = 720
My question why can't i say first digit can be chosen by 9 ways second digit can be chosen by 9 ways third digit can be chosen by 8 ways and we can arrange it in 3 ways, thus the answer is 9*9*8*3 Ah, I got your question. Yes, the easiest way would be as you pointed out: A. Numbers without restriction = 9^3; B. 3digit integers with all alike digits = 9 (111, 222, ..., 999); AB=9^39=720. Now, if we go your way it should be 9*8*7+9*8*3=720, where 9*8*7 is # of 3digit integers with all distinct digits and 9*8*3 is # of 3digit integers with two alike digits (XXY  9*8 and multiplied by 3!/2! to get different permutations). So, as there are two cases you cannot combine them in one formula and apply one factorial correction (3!/2!) to it. Hope it's clear. Sorry I don't understand both ways you mentioned. I dont know what is wrong with my way 9 ways for the first digit 9 ways for the second digit 8 ways for the last one => 9 x 9 x 8 = 648 ways. please explain! Many thanks We want to find the number of 3digit integers, without 0, in which none of the digits appear more than twice. So, the number should be XXY (2 digits are alike) or XYZ (all digits are distinct). Now, {all 3digit integers, without 0} = {3digit integers with all the same digits} + {3digit integers where 2 digits are alike} + {3digit integers with distinct digits}. Therefore, {3digit integers where 2 digits are alike} + {3digit integers with distinct digits} = {all 3digit integers, without 0}  {3digit integers with the same digits}. {3digit integers with 2 digits are alike} + {3digit integers with distinct digits} = 9^3  9. Hope it helps.
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Re: How many 3 digit integers can be chosen such that none of
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29 Nov 2014, 08:00
HuongShashi wrote: Bunuel wrote: Silver89 wrote: Thanks, In fact you can solve it in another way too which is number of ways without restriction = 9*9*9 = 729 number of ways that break restriction = 9*1*9 = 9 so the answer is 7299 = 720
My question why can't i say first digit can be chosen by 9 ways second digit can be chosen by 9 ways third digit can be chosen by 8 ways and we can arrange it in 3 ways, thus the answer is 9*9*8*3 Ah, I got your question. Yes, the easiest way would be as you pointed out: A. Numbers without restriction = 9^3; B. 3digit integers with all alike digits = 9 (111, 222, ..., 999); AB=9^39=720. Now, if we go your way it should be 9*8*7+9*8*3=720, where 9*8*7 is # of 3digit integers with all distinct digits and 9*8*3 is # of 3digit integers with two alike digits (XXY  9*8 and multiplied by 3!/2! to get different permutations). So, as there are two cases you cannot combine them in one formula and apply one factorial correction (3!/2!) to it. Hope it's clear. Sorry I don't understand both ways you mentioned. I dont know what is wrong with my way 9 ways for the first digit 9 ways for the second digit 8 ways for the last one => 9 x 9 x 8 = 648 ways. please explain! Many thanks As bunnel aptly said, it is quite simple to follow two steps : 1) 9*9*9 =729 ;this includes the no , which are repeated more than twice. 2) we have 9 no's excluding 0 ,whose all digits are same. Hence , total no =7299=720 Answer B



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Re: How many 3 digit integers can be chosen such that none of
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28 Mar 2015, 12:00
Direct with restriction: Let m = number of threedigit integers with one digit appearing twice m=9*1*8=72 However, there are C(3,2) ways that this condition can occur. M=Total Appearing Twice=m*C(3,2)=72*3=216
Let z=number of threedigit integers with each digit appearing once. z=9*8*7=504 There are C(3,3) ways that this condition can occur Z=Total Appearing Once=z*C(3,3)=504*1=504
Total satisfying the problem condition=M+Z=216+504=720
B



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Re: How many 3 digit integers can be chosen such that none of
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29 Mar 2015, 08:32
Silver89 wrote: How many 3 digit integers can be chosen such that none of the digits appear more than twice and none of the digits equal zero ?
A. 729 B. 720 C. 648 D. 640 E. 576
what is the wrong of my method which is : 3*(9*9*8) where first digit can be chosen by 9 ways second digit can be chosen by 9 ways third digit can be chosen by 8 ways and since it's anagram of ( SSD ) it can be arranged in 3 ways so the answer IMO is 3*(9*9*8)= 1944 which is wrong The simplest way to solve this is to find the numbers without any conditions applied and then from that subtract the numbers the satisfy the conditions. Total 3digit numbers without any digit being 0 = 9 * 9 * 9 = 729. Out of these only 111, 222, 333, ... , 999 violate the given condition that a digit cannot be repeated more than twice. Hence there are 9 such numbers. Hence 729  9 = 720. Hence option (B).  Optimus Prep's GMAT On Demand course for only $299 covers all verbal and quant. concepts in detail. Visit the following link to get your 7 days free trial account: http://www.optimusprep.com/gmatondemandcourse



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Re: How many 3 digit integers can be chosen such that none of
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30 Mar 2015, 02:40
Let's say this number is ABC.
Let's calculate how many numbers with two similar non zero digits and how many numbers with distinct non zero digits:
1. Two similar none zero digits: A has 9 possibilities, B has 1 possibility and C has 8 possibilities. So 9.1*8=72 different numbers. However, this configuration can be modeled in 3 ways, depending on the position of the similar numbers: XXY, XYX, YXX, therefore 72*3=216.
2. Distinct numbers: A has 9 possibilities, B has 8 possibilities and C has 7 possibilities, therfore: 9*8*7=504.
In total we have 504+216=720.



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How many 3 digit integers can be chosen such that none of
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08 Sep 2016, 20:37
How many 3 digit integers can be chosen such that none of the digits appear more than twice and none of the digits equal zero?
9/9*9/10*9/10=729/900 7299 triples=720 B. 720



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Re: How many 3 digit integers can be chosen such that none of
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11 Nov 2017, 03:32
I solved it by 9 * 9 * 8
First digit nine ways Second digit nine ways since we can repeat the number twice Third number 8
Why is it wrong?



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Re: How many 3 digit integers can be chosen such that none of
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11 Nov 2017, 05:10
santro789 wrote: I solved it by 9 * 9 * 8
First digit nine ways Second digit nine ways since we can repeat the number twice Third number 8
Why is it wrong? Please read carefully here: https://gmatclub.com/forum/howmany3d ... l#p1035258
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Re: How many 3 digit integers can be chosen such that none of
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04 Jul 2018, 03:11
TooLong150 wrote: Direct with restriction: Let m = number of threedigit integers with one digit appearing twice m=9*1*8=72 However, there are C(3,2) ways that this condition can occur. M=Total Appearing Twice=m*C(3,2)=72*3=216
Let z=number of threedigit integers with each digit appearing once. z=9*8*7=504 There are C(3,3) ways that this condition can occur Z=Total Appearing Once=z*C(3,3)=504*1=504
Total satisfying the problem condition=M+Z=216+504=720
B Can someone explain the logic behind "There are C(3,3) ways that this condition can occur". I've never seen this.



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How many 3 digit integers can be chosen such that none of
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23 Sep 2019, 08:00
Silver89 wrote: How many 3 digit integers can be chosen such that none of the digits appear more than twice and none of the digits equal zero ?
A. 729 B. 720 C. 648 D. 640 E. 576 SOLUTION 1 (CASES) \(digits={1,2,3,4,5,6,7,8,9}=9\) \([000:all.different]…9•8•7=72•7=490+14=504\) \([110:pair.different]…9•1•8•arrangements(3!/2!)=72•3=216\) \(total=504+216=720\) Answer (B) SOLUTION 2 (TOTALNOTCASE) \(digits={1,2,3,4,5,6,7,8,9}=9\) \([XXX:any.digit]…9•9•9=729\) \([333:triple]…9•1•1=9\) \(totalnot=7299=720\) Answer (B)




How many 3 digit integers can be chosen such that none of
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