Author 
Message 
TAGS:

Hide Tags

Manager
Status: Trying to survive
Joined: 29 Jun 2011
Posts: 170
GMAt Status: Quant section
Concentration: Finance, Real Estate
GMAT Date: 12302011
GPA: 3.2

How many 3 digit integers can be chosen such that none of [#permalink]
Show Tags
Updated on: 27 Jan 2012, 08:35
11
This post was BOOKMARKED
Question Stats:
55% (01:36) correct 45% (01:30) wrong based on 254 sessions
HideShow timer Statistics
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
Official Answer and Stats are available only to registered users. Register/ Login.
_________________
How can i lose my faith in life's fairness when i know that the dreams of those who sleep on the feathers are not more beautiful than the dreams of those who sleep on the ground?  Jubran Khaleel Jubran
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.



Manager
Status: Trying to survive
Joined: 29 Jun 2011
Posts: 170
GMAt Status: Quant section
Concentration: Finance, Real Estate
GMAT Date: 12302011
GPA: 3.2

Re: How many 3 digit integers can be chosen such that none of [#permalink]
Show Tags
27 Jan 2012, 08:40
1
This post received KUDOS
1
This post was BOOKMARKED
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
_________________
How can i lose my faith in life's fairness when i know that the dreams of those who sleep on the feathers are not more beautiful than the dreams of those who sleep on the ground?  Jubran Khaleel Jubran



Math Expert
Joined: 02 Sep 2009
Posts: 45430

How many 3 digit integers can be chosen such that none of [#permalink]
Show Tags
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.
_________________
New to the Math Forum? Please read this: Ultimate GMAT Quantitative Megathread  All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



Intern
Joined: 04 Mar 2014
Posts: 10
Concentration: Marketing, General Management

Re: How many 3 digit integers can be chosen such that none of [#permalink]
Show Tags
29 Nov 2014, 01:40
1
This post received KUDOS
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



Math Expert
Joined: 02 Sep 2009
Posts: 45430

Re: How many 3 digit integers can be chosen such that none of [#permalink]
Show Tags
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.
_________________
New to the Math Forum? Please read this: Ultimate GMAT Quantitative Megathread  All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



Manager
Joined: 21 Jan 2014
Posts: 62
WE: General Management (NonProfit and Government)

Re: How many 3 digit integers can be chosen such that none of [#permalink]
Show Tags
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



Manager
Joined: 10 Mar 2013
Posts: 243
GMAT 1: 620 Q44 V31 GMAT 2: 690 Q47 V37 GMAT 3: 610 Q47 V28 GMAT 4: 700 Q50 V34 GMAT 5: 700 Q49 V36 GMAT 6: 690 Q48 V35 GMAT 7: 750 Q49 V42 GMAT 8: 730 Q50 V39

Re: How many 3 digit integers can be chosen such that none of [#permalink]
Show Tags
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



SVP
Joined: 06 Nov 2014
Posts: 1888

Re: How many 3 digit integers can be chosen such that none of [#permalink]
Show Tags
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



Intern
Joined: 08 Sep 2014
Posts: 5

Re: How many 3 digit integers can be chosen such that none of [#permalink]
Show Tags
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.



Director
Joined: 07 Dec 2014
Posts: 999

How many 3 digit integers can be chosen such that none of [#permalink]
Show Tags
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



Manager
Joined: 30 Apr 2013
Posts: 90

Re: How many 3 digit integers can be chosen such that none of [#permalink]
Show Tags
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?



Math Expert
Joined: 02 Sep 2009
Posts: 45430

Re: How many 3 digit integers can be chosen such that none of [#permalink]
Show Tags
11 Nov 2017, 05:10




Re: How many 3 digit integers can be chosen such that none of
[#permalink]
11 Nov 2017, 05:10






