Notice that we are told that "Each number is to consist of four different digits from 0 to 9, inclusive, except that the first digit cannot be 0".

The first digit can take 9 values from 1 to 9 inclusive;
The second digit can also take 9 values (9 digits minus the one we used for the first digit plus 0);
The third digit can take 8 values;
The fourth digit can take 7 values.

Total = 9*9*8*7 = something with the units digit of 6.

Answer: B.

Hope it's clear.
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First digit (1000th place) can be any of 0 to 9 except 0, so it can be chosen in nine ways

Second digit (100th place) can be any of 0 to 9 except the one already chosen for 1000th place, so it can be chosen in nine ways

Third digit (10th place) can be any of 0 to 9 except the ones already chosen for 1000th place and 100th place, so it can be chosen in eight ways

Fourth digit (units place) can be any of 0 to 9 except the ones already chosen for 1000th place, 100th place and 10th place, so it can be chosen in seven ways

Total number of ways = 9*9*8*7 = 81*56. Only option with 6 in units place is B, so answer is B.

"Each number is to consist of four different digits from 0 to 9, inclusive, except that the first digit cannot be 0"
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Check Constructing Numbers, Codes and Passwords in our Special Questions Directory.
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1st digit can be anything from 1 to 9, So 9 possible digits
2nd can be anything from 0 to 9 but not the one used as 1st digit, so again 10- 1 = 9 possible values
3rd can be anything from 0 to 9 except the two digits used as 1st & 2nd digit, so 10-2 = 8 possible values
4th can be anything from 0 to 9 except the 3 digits used as 1st, 2nd and 3rd digit, so 10-3 = 7

Hence total possibilities = 9x9x8x7 = 4536

Hit kudos if you like the solution!!
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1. Ordering is important and there is no repetition, so this is an nPr problem
2. Is there a constraint?. The first digit cannot be zero
3. Number of ways the first digit can be selected is 9
4. Number of ways the second third and last digits can be ordered is 9P3
5. Total number of permutations is 9*9*8*7=4536
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Yes, that's correct.
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Immediate application of the Multiplicative Principle:

\(\begin{array}{*{20}{c}}
{\underline {{\text{not}}\,\,0} } \\
9
\end{array}\begin{array}{*{20}{c}}
{\underline {{\text{nr}}} } \\
9
\end{array}\begin{array}{*{20}{c}}
{\underline {{\text{nr}}} } \\
8
\end{array}\begin{array}{*{20}{c}}
{\underline {{\text{nr}}} } \\
7
\end{array}\,\,\,\,\,\mathop \Rightarrow \limits^{{\text{Multipl}}{\text{.}}\,{\text{Principle}}} \,\,\,\,? = {9^2} \cdot 8 \cdot 7\,\,\,\,\,\,\,\,\,\,\left[ {nr = {\text{no}}\,\,{\text{repetition}}} \right]\)

\(\left\langle ? \right\rangle = \left\langle {{9^2}} \right\rangle \cdot \left\langle {8 \cdot 7} \right\rangle = 1 \cdot 6 = 6\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left[ {\left\langle N \right\rangle = {\text{units}}\,\,{\text{digit}}\,\,{\text{of}}\,\,N} \right]\)


Just one alternative choice with unit´s digit equal to the correct one... we are done!


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

Regards,
Fabio.
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total digit =10
4 digits first cannot be 0
so 9*9*8*7 ; 4536
IMO B

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