How many positive five-digit integers contain the digit

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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

The OA is D.

Source: Manhattan Prep

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grouping 57

by GMATGuruNY » Thu Nov 01, 2018 9:27 am
I believe that the prompt above is outdated; regardless, it does not list a correct answer. The prompt below appears in Manhattan's 5 lb. Book of GRE Practice Problems:
How many times does the digit grouping "57" (in that order) appear in all of the five-digit positive integers? For instance, "57" appears once in 12,357, twice in 57,057, and does not appear in 24,675.

A. 279
B. 3,000
C. 3,500
D. 3,700
E. 4,000
Case 1: 57XXX
Number of options for the hundreds digit = 10. (Any digit 0-9.)
Number of options for the tens digit = 10. (Any digit 0-9.)
Number of options for the units digit = 10. (Any digit 0-9.)
To combine these options, we multiply:
10*10*10 = 1000.

Remaining cases: X57XX, XX57X, XXX57
Number of possible positions for the "57" grouping = 3. (The 3 cases above.)
Number of options for the leftmost digit = 9. (Any digit but 0.)
Number of options for the rightmost digit = 10. (Any digit 0-9.)
Number of options for the last remaining digit = 10. (Any digit 0-9.)
To combine these options, we multiply:
3*9*10*10 = 2700.

Total ways = Case 1 + Case 2 = 1000 + 2700 = 3700.

The correct answer is D.
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swerve wrote: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
Source: Manhattan Prep
The solution posted above is related to another problem (as mentioned).

Let´s solve the one originally proposed, in which there are double-counting´s to be dealt with!


$$?\,\,:\,\,5{\rm{ - digit}}\,\,{\rm{positive}}\,\,{\rm{integers}}\,\,{\rm{with}}\,\,57{\rm{ - block}}\left( {\rm{s}} \right)$$
$$\eqalign{
& \left( {\rm{1}} \right)\,\,\,\underline 5 \,\,\, \underline 7 \,\,\, \underline {} \,\,\, \underline {} \,\,\, \underline {} \,\,\,\,\, \to \,\,\,\,{10^3}\,\,{\rm{ways}} \,\, \cr
& \left( {\rm{2}} \right)\,\,\,\underline {{\rm{not}}\,0} \,\,\, \underline 5 \,\,\, \underline 7 \,\,\, \underline {} \,\,\, \underline {} \,\,\,\,\, \to \,\,\,\,9 \cdot {10^2}\,\,{\rm{ways}} \,\, \cr
& \left. \matrix{
\left( {\rm{3}} \right)\,\,\,\underline {{\rm{not}}\,0} \,\,\, \underline {} \,\,\, \underline 5 \,\,\, \underline 7 \,\,\, \underline {} \,\,\,\,\, \to \,\,\,\,9 \cdot {10^2}\,\,{\rm{ways}} \hfill \cr
\left( - \right)\,\,\,\underline {\rm{5}} \,\,\, \underline 7 \,\,\, \underline 5 \,\,\, \underline 7 \,\,\, \underline {} \,\,\,\,\, \to \,\,\,\,10\,\,{\rm{ways}} \hfill \cr} \right\}\,\,\,\, \to \,\,\,\,\,890\,\,{\rm{ways}} \,\, \cr
& \left. \matrix{
\left( {\rm{4}} \right)\,\,\,\underline {{\rm{not}}\,0} \,\,\, \underline {} \,\,\, \underline {} \,\,\, \underline 5 \,\,\, \underline 7 \,\,\,\,\, \to \,\,\,\,9 \cdot {10^2}\,\,{\rm{ways}} \hfill \cr
\left( - \right)\,\,\,\underline {{\rm{not}}\,0} \,\,\, \underline 5 \,\,\, \underline 7 \,\,\, \underline 5 \,\,\, \underline 7 \,\,\,\,\, \to \,\,\,\,9\,\,{\rm{ways}} \hfill \cr
\left( - \right)\,\,\,\underline 5 \,\,\, \underline 7 \,\,\, \underline {} \,\,\, \underline 5 \,\,\, \underline 7 \,\,\,\,\, \to \,\,\,\,10\,\,{\rm{ways}} \hfill \cr} \right\}\,\,\,\, \to \,\,\,\,\,881\,\,{\rm{ways}} \cr} $$
$$? = 1000 + 900 + 890 + 881 = 3671$$

The alternative choices are therefore all wrong.


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

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Fabio.
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