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Re: Of the threedigit positive integers whose three digits are all differ
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04 Nov 2017, 05:08
pra1785 wrote: ENGRTOMBA2018 wrote: Noxy416 wrote: Of the threedigit positive integers whose three digits are all different and nonzero, how many are odd integers greater than 700? (A) 84 (B) 91 (C) 100 (D) 105 (E) 243 What am i doing wrong in this question?
Number of options for the hundreds digit =3 Number of options for the Units digit =4 Number of options for the Units digit = 8
Total number of options 3 x 8 x 4 = 96 Please help Please follow posting guidelines (link in my signatures). Do make sure to mention the correct title and put your analyses either in the next post or under spoilers.As for the question, you need to find the number of combinations possible such that: 1. It is a 3 digit number > 700 (or between 701999, inclusive). 2. All digits are different and NON ZERO. 3. The numbers must be ODD > the last digit can be 1 of 1,3,5,7,9 Based on this, the numbers can be of the following 3 types: Type 1: 7AB Type 2: 8EF Type 3: 9CD For type 1 and type 3, be very careful that the digits must be different. So if it is 7AB, then B can NOT be 7. Similarly for type 3, 9CD, D can NOT be 9. There is no such restriction when you find numbers of type 2 (8EF). Number of combinations for type 1 : 1*7*4 = 28 Number of combinations for type 2: 1*7*5 = 35 Number of combinations for type 3 : 1*7*4 = 28 Thus, total numbers possible = 28+35+28 = 91. B is thus the correct answer. Hope this helps. Hi, I am a little confused with the explanation. For number of combinations for type 2: why is it 1*7* 5Don't we have to make sure that all numbers are different, non zero and odd in type 2 as well? What am I missing? Hi.. The last digit is odd and can be any of the five digits 1,3,4,7,9 In type 2, the Hundreds digit is 8, so units digit can be any of 5 odd digits. But in other type, the Hundreds digit is an odd integer7 or 9, so units digit can be any of the remaining 4 as all digits are different
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Re: Of the threedigit positive integers whose three digits are all differ
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04 Nov 2017, 08:31
I am confused. 7AB 8BC 9CD.
Considering 7AB. 7 is fixed and is odd. Thus B will have 4 possibilities. But I think A has 8 possibilities after we fill one value with A. Let say we put 1 for B. A can have 0,2,3,4,5,6,8,9 i.e 8 possibilities. So for 1st case its 1*8*4=32 2nd case is 1*8*5=40 3rd case is same as 1st i.e 1*8*4=32 =>32+40+32=104. Please tell me where I am going wrong.



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Re: Of the threedigit positive integers whose three digits are all differ
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04 Nov 2017, 08:36
Utkarsh KOhli wrote: I am confused. 7AB 8BC 9CD.
Considering 7AB. 7 is fixed and is odd. Thus B will have 4 possibilities. But I think A has 8 possibilities after we fill one value with A. Let say we put 1 for B. A can have 0,2,3,4,5,6,8,9 i.e 8 possibilities. So for 1st case its 1*8*4=32 2nd case is 1*8*5=40 3rd case is same as 1st i.e 1*8*4=32 =>32+40+32=104. Please tell me where I am going wrong. hi.. be careful while reading the statements. many times we all tend to overlook finer points.. same case here.. it says three digits are different and NON ZERO.. so coloured portion above will be 7 possiblities
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Re: Of the threedigit positive integers whose three digits are all differ
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04 Nov 2017, 10:08
So we have 3 digits numbers starting with 7 , 8 and 9. No repetitive digits, no zeros, numbers should be odd so ending with odd digit 7AB, 8AB, 9AB where A even number (excluding 0) and B is odd number (excluding repetition) even numbers are 2, 4, 6, 8 odd numbers are 1, 3, 5, ,7 ,9
So for 7AB 1*4*4 (as we have to exclude 7 from B) = 16 So for 8AB 1*3*5 (as we have to exclude 8 from A) = 15 So for 9AB 1*4*4 (as we have to exclude 9 from B) = 16
Now we can have 7BB type combinations well, where BB are both odd numbers So for 7BB 1*4*3 (as we have to exclude 7 from B) = 12 So for 8BB 1*5*4 = 20 So for 9BB 1*4*3 (as we have to exclude 9 from B) = 12
Total 16+15+16+12+20+12=91



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Re: Of the threedigit positive integers whose three digits are all differ
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04 Nov 2017, 10:26
chetan2u wrote: Utkarsh KOhli wrote: I am confused. 7AB 8BC 9CD.
Considering 7AB. 7 is fixed and is odd. Thus B will have 4 possibilities. But I think A has 8 possibilities after we fill one value with A. Let say we put 1 for B. A can have 0,2,3,4,5,6,8,9 i.e 8 possibilities. So for 1st case its 1*8*4=32 2nd case is 1*8*5=40 3rd case is same as 1st i.e 1*8*4=32 =>32+40+32=104. Please tell me where I am going wrong. hi.. be careful while reading the statements. many times we all tend to overlook finer points.. same case here.. it says three digits are different and NON ZERO.. so coloured portion above will be 7 possiblities Hi Chetan. Thank you for pointing it out. Kudos to you.



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Re: Of the threedigit positive integers whose three digits are all differ
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31 Jan 2018, 20:51
Hi All, You'll likely find it easiest to handle this calculation in 'pieces' This question lays out the following restrictions: 1) 3digit numbers greater than 700 2) All digits are NON0 and DIFFERENT 3) The number must be ODD. Let's start with the 700s... 1st digit is 7 = 1 option 3rd digit must be ODD, but NOT 7 = 4 options 2nd digit must differ from the other two digits and not be 0 = 7 options (1)(4)(7) = 28 numbers in the 700s Now, the 800s... 1st digit is 8 = 1 option 3rd digit must be ODD = 5 options 2nd digit must differ from the other two digits and not be 0 = 7 options (1)(5)(7) = 35 numbers in the 800s Finally, the 900s; the math here works the same as the 700s... 1st digit is 9 = 1 option 3rd digit must be ODD, but NOT 9 = 4 options 2nd digit must differ from the other two digits and not be 0 = 7 options (1)(4)(7) = 28 numbers in the 900s 28+35+28 = 91 Final Answer: GMAT assassins aren't born, they're made, Rich
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Re: Of the threedigit positive integers whose three digits are all differ
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18 Oct 2018, 21:54
TypeI: 7AB TypeII: 8CD TypeIII 9EF
TypeI = 1* A * B( Odd=1,3,5,7,9. Minus 7= 1,3,5,9= 4) So, 1 * A * 7 (A is not 7 or B= 1,2,3,4,5,6,7,8,9= 7 and B= 7) Now TypeI is 1*7*4
TypeII = 1 * C * D (D= odd= 1,3,5,7,9 = 5/ C= 1,2,3,4,5,6,7,8,9 minus 8 and C= 7) So, 1* 7* 5
TypeIII is the same as TypeI
TypeI: 1*7*4= 28 TypeII: 1*7*5=35 TypeIII 1*7*4=28 Answer is 91




Re: Of the threedigit positive integers whose three digits are all differ
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