How many positive five-digit integers contain the digit grouping 57

Question

How many positive five-digit 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

(A) 279

(B) 3,000

(C) 3,500

(D) 3671

(E) 3,700

KarishmaB · Accepted Answer

Responding to a pm: Reminds me of questions where two people must sit together. I first make 57 take a position of the 4 available ones (1st, 2nd, 3rd or 4th) Now there are two cases: Case 1 : 57 takes the first (and second) spot 57 _ _ _ We have 3 spots and 10 options for each so 10 *10 *10 = 1000 numbers Case 2 : 57 takes the second, third or fourth spots _ 57 _ _ ; _ _ 57 _ ; _ _ _ 57 For the first spot, we have only 9 options (0 cannot be there since we need 5 digit numbers) For the remaining 2 spots, we have 10 options each. So total 900*3 = 2700 numbers But we are double counting those cases in which 57 appears twice (It cannot appear thrice). e.g. the number 57573 is counted in case 1 as well as case 2. So we need to remove it once so that it is counted only once. How many such cases are there? It is now a sub question: How many 5 digit numbers have two pairs of 57 in them? We can place two 57s in the spots in the following cases: 1 and 3: 5757_ (10 such numbers) 1 and 4: 57 _ 57 (10 such numbers) 2 and 4: _ 5757 (9 such numbers) So from 3700, we remove 29 numbers to get such 3671 numbers. Answer: 3671 numbers Check this video: https://youtu.be/LFnLKx06EMU

VeritasPrepBrian · Answer

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 ten-thousands digit where this would happen, since you can't start with 0)

So that's the 29 options that are double counted.

SonalSinha803 · Answer

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.

Posted from my mobile device

u1983 · Answer

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 5-digit 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 .

u1983 · Answer

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 5-digit 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 :  3700-29 = 3671 ............................................ Might be the desired answer  Please correct me if I am wrong

Helium · Answer

"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, XXXY

so 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

fskilnik · Answer

?\,\,:\,\,5{m{ - digit}}\,\,{m{positive}}\,\,{m{integers}}\,\,{m{with}}\,\,57{m{ - block}}\left( {m{s}} ight)

There are double-counting´s to be dealt with!

\eqalign{
 & \left( {m{1}} ight)\,\,\,\underline 5 \,\,\, \underline 7 \,\,\, \underline {} \,\,\, \underline {} \,\,\, \underline {} \,\,\,\,\, 	o \,\,\,\,{10^3}\,\,{m{ways}} \,\, \cr 
 & \left( {m{2}} ight)\,\,\,\underline {{m{not}}\,0} \,\,\, \underline 5 \,\,\, \underline 7 \,\,\, \underline {} \,\,\, \underline {} \,\,\,\,\, 	o \,\,\,\,9 \cdot {10^2}\,\,{m{ways}} \,\, \cr 
 & \left. \matrix{
 \left( {m{3}} ight)\,\,\,\underline {{m{not}}\,0} \,\,\, \underline {} \,\,\, \underline 5 \,\,\, \underline 7 \,\,\, \underline {} \,\,\,\,\, 	o \,\,\,\,9 \cdot {10^2}\,\,{m{ways}} \hfill \cr 
 \left( - ight)\,\,\,\underline {m{5}} \,\,\, \underline 7 \,\,\, \underline 5 \,\,\, \underline 7 \,\,\, \underline {} \,\,\,\,\, 	o \,\,\,\,10\,\,{m{ways}} \hfill \cr} ight\}\,\,\,\, 	o \,\,\,\,\,890\,\,{m{ways}} \,\, \cr 
 & \left. \matrix{
 \left( {m{4}} ight)\,\,\,\underline {{m{not}}\,0} \,\,\, \underline {} \,\,\, \underline {} \,\,\, \underline 5 \,\,\, \underline 7 \,\,\,\,\, 	o \,\,\,\,9 \cdot {10^2}\,\,{m{ways}} \hfill \cr 
 \left( - ight)\,\,\,\underline {{m{not}}\,0} \,\,\, \underline 5 \,\,\, \underline 7 \,\,\, \underline 5 \,\,\, \underline 7 \,\,\,\,\, 	o \,\,\,\,9\,\,{m{ways}} \hfill \cr 
 \left( - ight)\,\,\,\underline 5 \,\,\, \underline 7 \,\,\, \underline {} \,\,\, \underline 5 \,\,\, \underline 7 \,\,\,\,\, 	o \,\,\,\,10\,\,{m{ways}} \hfill \cr} ight\}\,\,\,\, 	o \,\,\,\,\,881\,\,{m{ways}} \cr}

? = 1000 + 900 + 890 + 881 = 3671

This solution follows the notations and rationale taught in the GMATH method.

Regards, 
Fabio.

Aboyhasnoname · Answer

Bunuel  Could you help here..Some explanations say overcounting bringing the final answer to 3671......I am a little confused..... Maybe something like this

Case 1... 
5 7 _ _ _ = 10 x 10 x 10 Numbers.... 
in Another Case...

_ _ 5 7 _ = 9 X 10 x 1 x 1 x 10 = 900 Numbers..
But this includes cases ..Where the number is 57571 or 57572...
The double counting is happening here?.....

Bunuel  if you could help?....
Thanks

Bunuel · Answer

Yes. Check correct solution here: https://gmatclub.com/forum/how-many-pos ... l#p3257633

How many positive five-digit integers contain the digit grouping 57

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How many positive five-digit integers contain the digit grouping 57

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