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A 3digit positive integer consists of non zero digits. If
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15 Aug 2011, 01:22
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A 3digit positive integer consists of non zero digits. If exactly two of the digits are the same, how many such integers are possible? (A) 72 (B) 144 (C) 216 (D) 283 (E) 300
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Re: Probability  Exactly Question!
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15 Aug 2011, 01:51
DeeptiM wrote: A 3digit positive integer consists of non zero digits. If exactly two of the digits are the same, how many such integers are possible? (A) 72 (B) 144 (C) 216 (D) 283 (E) 300 here is the answer in pure probability terms. let the 3 digit number be XYZ now none of them contain zero > they can vary from 1  9 so lets take the value of X, X can be from 1  9, hence X can be selected in 9 ways since Y =X, after selecting X you can select Y in 1 way only Z can be an selected from 1 9 (minus 1 because it cannot be the same as X) in exactly 8 ways hence total number of ways of selecting the number is 9 * 8 = 72. BUT we are asked for all possible numbers that can be formed. So XYZ ( or XXZ in this case) can be moved around in 3 ways XXZ XZX ZXX hence the total number of numbers is 72 *3 = 216



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Re: Probability  Exactly Question!
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15 Aug 2011, 02:00
Ques says..non zero digits..y didnt u consider the negative values?



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Re: Probability  Exactly Question!
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15 Aug 2011, 02:04
the first few words in ur ques says "A 3digit positive integer " although im not Einstein i dont think we can have a postive 3 digit number with negative digits



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Re: Probability  Exactly Question!
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15 Aug 2011, 02:07
Gosh, I cant stop laughing on my sillyness...lol



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Re: Probability  Exactly Question!
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15 Aug 2011, 02:19
we just have to make sure we dont miss such things on the real test... or we just butchered our test and you might watch your months of preps flowing down the drain!!



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Re: Probability  Exactly Question!
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15 Aug 2011, 02:23
i knwww and the worst part is I keep doing these mistakes..boo hoo..
thanks for all the help Vikas..



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Re: Probability  Exactly Question!
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15 Aug 2011, 02:28
DeeptiM wrote: A 3digit positive integer consists of non zero digits. If exactly two of the digits are the same, how many such integers are possible? (A) 72 (B) 144 (C) 216 (D) 283 (E) 300 Same as viks4gmat: \(C^{9}_{2}*2*\frac{3!}{2!}=36*2*3=216\) \(C^9_2:\) Number of ways we can select 2 digits out of 9 {1,2,3,4,5,6,7,8,9} We can select 1,2 OR 4,9 But, we need to multiply each of these selections by 2. Why? For 8,9 8,8,9 8,9,9 Are two possibilities. First, in which we choose 2 8's and second, in which we choose 2 9's. If we have; XXY It can be uniquely arranged in = 3!/2! ways 3 terms: X,X,Y=3!(numerator) 2 repeats: X,X=2!(denominator) So, if we select 889  889 898 988 Total=3 ways Ans: "C"
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Re: Probability  Exactly Question!
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15 Aug 2011, 02:33
DeeptiM wrote: Ques says..non zero digits..y didnt u consider the negative values? If GMAT uses the word "digit", it is always nonnegative, single digit integers. There are only 10 possible digits: {0,1,2,3,4,5,6,7,8,9} So, nonzero digits {1,2,3,4,5,6,7,8,9} A real number integers, rational/irrational numbers, decimals, etc. can be negative, positive or zero.
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Re: A 3digit positive integer consists of non zero digits. If
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23 Dec 2015, 04:52
DeeptiM wrote: A 3digit positive integer consists of non zero digits. If exactly two of the digits are the same, how many such integers are possible? (A) 72 (B) 144 (C) 216 (D) 283 (E) 300 i took a different approach but missed something total possibilities : 9x9x9 = 729 no two digits are the same : 9x8x7 = 504 exactly two = 225 ? what did i miss?



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Re: A 3digit positive integer consists of non zero digits. If
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23 Dec 2015, 05:01
jimwild wrote: DeeptiM wrote: A 3digit positive integer consists of non zero digits. If exactly two of the digits are the same, how many such integers are possible? (A) 72 (B) 144 (C) 216 (D) 283 (E) 300 i took a different approach but missed something total possibilities : 9x9x9 = 729 no two digits are the same : 9x8x7 = 504 exactly two = 225 ? what did i miss? your answer 225 consists of all three digit same .. iy will be 9*1*1=9.. so actual answer=2259=216.. hope it helps
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Re: A 3digit positive integer consists of non zero digits. If
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26 May 2017, 22:10
DeeptiM wrote: A 3digit positive integer consists of non zero digits. If exactly two of the digits are the same, how many such integers are possible? (A) 72 (B) 144 (C) 216 (D) 283 (E) 300 1. When units and tens digits are the same. There are 9 such cases. For each, hundreds digit can have 8 values for a total of 72 2. Similarly, units and hundreds and also tens and hundreds can have the same values, for a total of 72*3=216
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Re: A 3digit positive integer consists of non zero digits. If
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23 Oct 2018, 04:55
DeeptiM wrote: A 3digit positive integer consists of non zero digits. If exactly two of the digits are the same, how many such integers are possible? (A) 72 (B) 144 (C) 216 (D) 283 (E) 300 I thought of the question this way.. I first thought of making 1 as the number that appears twice. There are three ways : 11x, 1x1 and x11 where x is a number from 2 to 9. We find that there 24 possible ways= 3 ways * 8 (because any number from 2 to 9 can be there in the space of x) = 24. Now multiply 24 into 9 because now any number from 2 to 9 can be the two common digits. 24 * 9 = 216.



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Re: A 3digit positive integer consists of non zero digits. If
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23 Oct 2018, 05:08
DeeptiM wrote: A 3digit positive integer consists of non zero digits. If exactly two of the digits are the same, how many such integers are possible? (A) 72 (B) 144 (C) 216 (D) 283 (E) 300 We need 3 digit integer with non zero digits (so only 1 to 9 are allowed). Since 2 of the digits are the same, the numbers are of the form XXY or XYX or YXX (so 3 forms). You can select X in 9 ways and Y in 8 ways. Total integers possible = 9*8*3 = 216 Answer (C)
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Re: A 3digit positive integer consists of non zero digits. If
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16 May 2019, 19:59
DeeptiM wrote: A 3digit positive integer consists of non zero digits. If exactly two of the digits are the same, how many such integers are possible? (A) 72 (B) 144 (C) 216 (D) 283 (E) 300 The 3 digits are nonzero, and they consist of two distinct digits. The number of ways to choose 2 digits from 9 nonzero digits is 9C2 = (9 x 8)/2 = 36. Now for each pair of digits chosen, for example, 1 and 2, we could have: 112, 121, 211, 221, 212, and 122. Therefore, there are 36 x 6 = 216 such integers. Answer: C
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Re: A 3digit positive integer consists of non zero digits. If
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