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Of the threedigit positive integers that have no digits [#permalink]
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31 Jan 2007, 06:12
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Of the threedigit positive integers that have no digits equal to zero, how many have two digits that are equal to each other and the remaining digit different from the other two? A. 24 B. 36 C. 72 D. 144 E. 216
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Re: PS  Digits (Very Tough one) [#permalink]
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31 Jan 2007, 07:21
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TOUGH GUY wrote: Of the threedigit positive integers that have no digits equal to zero, how many have two digits that are equal to each other and the remaining digit different from the other two?
Assume: a,b,c is the digit and a,b,c not= 0
Thus a,b,c could be 1, 2, 3, 4, 5, 6, 7, 8, and 9
Count the possible way to get three digit number abc.
a = b, and c must be different from a, b
Thus, there are 3 possible ways of digit arrangement: aac, aca, caa
Case I: aac
=> (digit 1st) x (digit 2nd) x (digit 3rd)
=> 9 x 1 x 8 {pick any number from group = 9 possible ways} x {pick number the same as the first pick = 1 way} x {pick any number from the rest = 8 possible ways}
= 9 x 1 x 8 = 72 possible ways
Case II: aca
=> same as case I you have 72 possible ways
Case III: caa
=> same as case I you have 72 possible ways
total of this set of number = 72 + 72 + 72 = 216
E) is the answer
(I got this question the first time 72 but I found that is question needs a little more work to do )



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Re: PS  Digits (Very Tough one) [#permalink]
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31 Jan 2007, 22:49
TOUGH GUY wrote: Of the threedigit positive integers that have no digits equal to zero, how many have two digits that are equal to each other and the remaining digit different from the other two?
A. 24 B. 36 C. 72 D. 144 E. 216
threedigit positive integer where two digits that are equal to each other and the remaining digit different from the other two
A B C three digits
possible combinations
AAC
ACA
CAA
11(2,3,4,5,6,7,8,9) total 8 combinations
1((2,3,4,5,6,7,8,9)1 total 8 combinations
(2,3,4,5,6,7,8,9)1,1 total 8 combinations
3*8=24
and so on
we have 9 digits with 24 possible comb for each
total 9*24=216
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Zero can't be a digit so the not equal can be selected 9 ways and equal can be selected 8 ways or vice versa.
If Number is XXY then there are 8*9=72 possibilities
If Number is XYX then there are 8*9 =72 possibilities
If Number is YXX then there are 8*9 =72 possibilities
Total 216
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Re: Of the threedigit positive integers that have no digits [#permalink]
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27 Jan 2012, 00:06



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Re: Of the threedigit positive integers that have no digits [#permalink]
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27 Jan 2012, 00:25
Baten80 wrote: many different ways to solve the problem. Is there any unique/good/universal approach that can be applied to other similar problems. I don' think that there is an unique or universal approach to such kind of problems. You should choose the one you are more comfortable with. As for the above problem. One can also use combination approach. Of the threedigit positive integers that have no digits equal to zero, how many have two digits that are equal to each other and the remaining digit different from the other two? A. 24 B. 36 C. 72 D. 144 E. 216 We have a three digit integer: XXY. Choosing the digit for X  9 ways; Choosing the digit for Y  8 ways; # of permutations of 3 digits in XXY  3!/2! (as 2 X's are identical); Total: 9*8*3!/2!=216. Answer: E. Hope it helps.
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Re: Of the threedigit positive integers that have no digits [#permalink]
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27 Jan 2012, 05:13
good question... good explanation too..
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Re: Of the threedigit positive integers that have no digits [#permalink]
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10 Feb 2014, 06:02
Bunuel wrote: Baten80 wrote: many different ways to solve the problem. Is there any unique/good/universal approach that can be applied to other similar problems. I don' think that there is an unique or universal approach to such kind of problems. You should choose the one you are more comfortable with. As for the above problem. One can also use combination approach. Of the threedigit positive integers that have no digits equal to zero, how many have two digits that are equal to each other and the remaining digit different from the other two? A. 24 B. 36 C. 72 D. 144 E. 216 We have a three digit integer: XXY. Choosing the digit for X  9 ways; Choosing the digit for Y  8 ways; # of permutations of 3 digits in XXY  3!/2! (as 2 X's are identical); Total: 9*8*3!/2!=216. Answer: E. Hope it helps. Is it possible to solve the inverse? Say TOTAL  (Cases were ALL are equal )  (Cases were ALL are different)? I'm trying to take a stab at it but I'm not able to come with the answer Is it 9^3  (9*3)  (9*8*7) ? Thanks Cheers J



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Re: Of the threedigit positive integers that have no digits [#permalink]
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10 Feb 2014, 06:11
jlgdr wrote: Bunuel wrote: Baten80 wrote: many different ways to solve the problem. Is there any unique/good/universal approach that can be applied to other similar problems. I don' think that there is an unique or universal approach to such kind of problems. You should choose the one you are more comfortable with. As for the above problem. One can also use combination approach. Of the threedigit positive integers that have no digits equal to zero, how many have two digits that are equal to each other and the remaining digit different from the other two? A. 24 B. 36 C. 72 D. 144 E. 216 We have a three digit integer: XXY. Choosing the digit for X  9 ways; Choosing the digit for Y  8 ways; # of permutations of 3 digits in XXY  3!/2! (as 2 X's are identical); Total: 9*8*3!/2!=216. Answer: E. Hope it helps. Is it possible to solve the inverse? Say TOTAL  (Cases were ALL are equal )  (Cases were ALL are different)? I'm trying to take a stab at it but I'm not able to come with the answer Is it 9^3  (9*3)  (9*8*7) ? Thanks Cheers J Almost nailed it! The # of threedigit integers that have all digits equal is just 9, not 9*3: 111, 222, ... 999. So, {Total}  {all equal}  {all distinct} = (9*9*9)  9  (9*8*7) = 216. Hope it's clear.
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Re: Of the threedigit positive integers that have no digits [#permalink]
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24 Sep 2016, 05:08
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Digits can be in the form ABB, BBA, BAB For ABB  9 options for A(19) and 8 options for B(29), since taking B as 1 will give 111 which is not required by question hence no. of combinations possible 9*8 = 72 similarly for BBA 9*8=72 and BAB  9*8 =72 total = 72+72+72 = 216 E is the answer



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Re: Of the threedigit positive integers that have no digits [#permalink]
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16 Oct 2016, 08:57
We want a number like : XXY, X and Y being digits from 1 to 9. To determine the number of possibilities, we can use the fundamental counting principle.
1/ Number of possibilities of X : 9 2/ Number of possibilities of Y : 8 3/ Number of ways to arranges XXY : 3!/2! = 3
So, 9*8*3 = 216



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Re: Of the threedigit positive integers that have no digits [#permalink]
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16 Oct 2016, 09:28
TOUGH GUY wrote: Of the threedigit positive integers that have no digits equal to zero, how many have two digits that are equal to each other and the remaining digit different from the other two?
A. 24 B. 36 C. 72 D. 144 E. 216 Let, here are three places that we have to fill by digits and two of them must be filled by same digit _ _ _There are three cases XXY ===> Total ways = 9*8 = 72 Numbers (X can be chosen 9 ways 19, and Y in another 8 ways) XYX ===> Total ways = 9*8 = 72 Numbers (X can be chosen 9 ways 19, and Y in another 8 ways) YXX ===> Total ways = 9*8 = 72 Numbers (Y can be chosen 9 ways 19, and Y in another 8 ways) Total Numbers = 3*72 = 216 Answer: Option E
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Re: Of the threedigit positive integers that have no digits [#permalink]
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15 Dec 2017, 14:15
Hi!
Can someone explain why the answer can't work using combination? For this example, I'm looking for the number of variations of AAB. I did 9 pick 2 (since you're using two digits and there's a total of 9 digits you can pick from). Then I multiplied 9 pick 2 (36) by 3!/2! which equals 108.



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Re: Of the threedigit positive integers that have no digits [#permalink]
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16 Dec 2017, 00:38
GMATKAY22 wrote: Hi!
Can someone explain why the answer can't work using combination? For this example, I'm looking for the number of variations of AAB. I did 9 pick 2 (since you're using two digits and there's a total of 9 digits you can pick from). Then I multiplied 9 pick 2 (36) by 3!/2! which equals 108. 9C2 = 36 is the number of pairs of different digits from 9 digits: (1, 2); (1, 3); (1, 4); ...; (8, 9). Now, one pair, say (1, 2), can give you TWO types of numbers: with two 1 and one 2 and two 2's and one 1: 112 (which in turn gives 112, 121, 211) and 221 (which in turn gives 221, 212, 122). So, you should multiply 9C2*3!/2! further by 2: 9C2*3!/2!*2 = 216. Hope it's clear.
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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.
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Re: Of the threedigit positive integers that have no digits
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