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How many positive fivedigit integers contain the digit grouping “57”
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23 May 2018, 06:59
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How many positive fivedigit integers contain the digit grouping “57” (in that order) at least once? For instance 30,457 and 20,574 are two such integers to include, but 30,475 and 20,754 do not meet the restrictions. (A) 279 (B) 3,000 (C) 3,500 (D) 3,700 (E) 4,000
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Re: How many positive fivedigit integers contain the digit grouping “57”
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23 May 2018, 07:06
The possibilities are 57*** , *57**, **57* and ***57.
Thus, the number of ways of doing so are  10 * 10 * 10 + 3* 10 * 10 * 9 = 3700.
Thus, imo D.
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Re: How many positive fivedigit integers contain the digit grouping “57”
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23 May 2018, 07:21
Bunuel wrote: How many positive fivedigit integers contain the digit grouping “57” (in that order) at least once? For instance 30,457 and 20,574 are two such integers to include, but 30,475 and 20,754 do not meet the restrictions.
(A) 279
(B) 3,000
(C) 3,500
(D) 3,700
(E) 4,000 From The Picture 01 : For case 01 : 3rd position can be filled in 10 ways ( available numbers 0,1,2,...9)... Similarly 4th position in 10 ways & 5th in 10 ways as well ==>Total cases= \(10*10*10 = 1000\) For case 02 : 1st position can be filled in 9 ways ( if 0, the number is not 5digit any more)... 4th position in 10 ways & 5th in 10 ways as well ==>Total cases= \(9*10*10 = 900\) For case 03 : 1st position can be filled in 9 ways , 2nd position in 10 ways & 5th in 10 ways as well ==>Total cases= \(9*10*10 = 900\) For case 04 : 1st position can be filled in 9 ways ,2nd position in 10 ways & 3rd in 10 ways as well ==>Total cases= \(9*10*10 = 900\) Hence , the total cases are \(1000 + 900 + 900 +900 = 3700\)........ Hence I would go for option D.
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Re: How many positive fivedigit integers contain the digit grouping “57”
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23 May 2018, 07:36
u1983 wrote: Bunuel wrote: How many positive fivedigit integers contain the digit grouping “57” (in that order) at least once? For instance 30,457 and 20,574 are two such integers to include, but 30,475 and 20,754 do not meet the restrictions.
(A) 279
(B) 3,000
(C) 3,500
(D) 3,700
(E) 4,000 From The Picture 01 : For case 01 : 3rd position can be filled in 10 ways ( available numbers 0,1,2,...9)... Similarly 4th position in 10 ways & 5th in 10 ways as well ==>Total cases= \(10*10*10 = 1000\) For case 02 : 1st position can be filled in 9 ways ( if 0, the number is not 5digit any more)... 4th position in 10 ways & 5th in 10 ways as well ==>Total cases= \(9*10*10 = 900\) For case 03 : 1st position can be filled in 9 ways , 2nd position in 10 ways & 5th in 10 ways as well ==>Total cases= \(9*10*10 = 900\) For case 04 : 1st position can be filled in 9 ways ,2nd position in 10 ways & 3rd in 10 ways as well ==>Total cases= \(9*10*10 = 900\) Hence , the total cases are \(1000 + 900 + 900 +900 = 3700\)........ Hence I would go for option D. Now the issue here is I have not considered a lot of cases . And thus I believe I am wrong . Experts please help. From The Picture 'Cases Open for Discussion : For case 01 : 3rd position can be filled in 10 ways ( available numbers 0,1,2,...9)... Similarly 4th position in 10 ways & 5th in 10 ways as well ==>Total cases= \(10*10*10 = 1000\).............. from here we need to exclude the cases where 57 at 3rd & 4th and 57 at 4th & 5th as these are to be considered at case 03 & case 04. A. 57 at 3rd & 4th while 57 at 1st & 2nd = 10 ( 5th position can be filled with 0,1,....9 ==>10 ways) B. 57 at 4th & 5th while 57 at 1st & 2nd = 10 ( 3rd position can be filled with 0,1,....9 ==>10 ways) For case 02 : 1st position can be filled in 9 ways ( if 0, the number is not 5digit any more)... 4th position in 10 ways & 5th in 10 ways as well ==>Total cases= \(9*10*10 = 900\).............. from here we need to exclude the case where 57 at 4th & 5th as these are to be considered at case 04. C. 57 at 4th & 5th while 57 at 2nd & 3rd = 9 ( 1st position can be filled with 1,....9 ==>9 ways) For case 03 : 1st position can be filled in 9 ways , 2nd position in 10 ways & 5th in 10 ways as well ==>Total cases= \(9*10*10 = 900\) For case 04 : 1st position can be filled in 9 ways ,2nd position in 10 ways & 3rd in 10 ways as well ==>Total cases= \(9*10*10 = 900\) Hence , the total overlapping cases are \(1000 + 900 + 900 +900 = 3700\) Hence , the total overlapped cases are \(A + B + C = 10 + 10 + 9 =29\) Hence total Unique cases : \(370029 = 3671\)............................................ Might be the desired answer Please correct me if I am wrong
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Re: How many positive fivedigit integers contain the digit grouping “57”
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24 May 2018, 23:56
Bunuel wrote: How many positive fivedigit integers contain the digit grouping “57” (in that order) at least once? For instance 30,457 and 20,574 are two such integers to include, but 30,475 and 20,754 do not meet the restrictions.
(A) 279
(B) 3,000
(C) 3,500
(D) 3,700
(E) 4,000 Possible combinations, 1) _, _, _, 57: first position can only be filled with numbers 1 to 9, 2nd with 0 to 9, similarly 3rd with 0 to 9, total arrangements  9 X 10 X 10 = 900 2) _, _, 57, _: first position can only be filled with numbers 1 to 9, 2nd with 0 to 9, similarly 5th with 0 to 9, total arrangements  9 X 10 X 10 = 900 3) _, 57, _, _: first position can only be filled with numbers 1 to 9, 4th with 0 to 9, similarly 5th with 0 to 9, total arrangements  9 X 10 X 10 = 900 4)57, _, _, _: 3rd position can be filled with numbers 0 to 9, 4th with 0 to 9, similarly 5th with 0 to 9, total arrangements  10 X 10 X 10 = 1000 Total possible arrangements = (3x900) + 1000 = 3700



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Re: How many positive fivedigit integers contain the digit grouping “57”
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25 May 2018, 15:38
I absolutely love this problem, but I agree with u1983 that the OA double counts some options (and you nailed the way to account for that and get to 3671. Visually, we have to subtract out: 57X57 (there are 10 values of X as a hundreds digit where this would happen) 5757X (there are 10 values of X as a units digit where this would happen) X5757 (there are 9 values of X as a tenthousands digit where this would happen, since you can't start with 0) So that's the 29 options that are double counted.
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Re: How many positive fivedigit integers contain the digit grouping “57”
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21 Aug 2018, 08:00
Bunuel wrote: How many positive fivedigit integers contain the digit grouping “57” (in that order) at least once? For instance 30,457 and 20,574 are two such integers to include, but 30,475 and 20,754 do not meet the restrictions.
(A) 279
(B) 3,000
(C) 3,500
(D) 3,700
(E) 4,000 "57" need to be treated as a group.So We can consider that we require 4 digit number where one digit is group "57". Say Y represents that digit > Y = "57"Required number XXXY. First digit can be selected in 10 ways. Same goes for 2 and 3 digit selection of required 4 digit number and digit Y can take 4 places. i.e. YXXX , XYXX, XXYX, XXXYso total 10*10*10*4 = 4000 such numbers.But if the first digit is 0 then we don't have 5 digit number ( 4 digit number as per our working). So let's subtract such cases. 0XXY or 0XYX or 0XXY 10 selections each for 2 X's and Y can take 3 places. So total 10*10*3 = 300 cases.Required answer = 4000  300 = 3700. Option  D
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Re: How many positive fivedigit integers contain the digit grouping “57” &nbs
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21 Aug 2018, 08:00






