A company plans to assign identification numbers to its empl

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.

Hi Bunnel,

I took 9*10*10*10=9000 i thought he didn't mention any thing such as repetition not allowed or not allowed may i know if i misread the question?

May i know where i went wrong?

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

Four DIFFERENT digits. 9*9*8*7.

Only answer with units digit 6 is B.

Hope this helps
Cheers
J

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The first thing to note is that the possibility of choices is 0-9- which actually is ten numbers if you count [0,1,2,3,4,5,6,7,8,9] yet the first number of the digit cannot have 0 so there is a pool of 9 choices to choose from for the first number

9 x 9 x 8 x 7 =
4536

Thus B

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!!

Since we have to form 4 digit number with 1st digit non-zero and all different digits.
It can be formed in following way
1st digit -> 9 ways (1 to 9)
2nd digit -> 9 ways (0 to 9 excluding 1st digit)
3rd digit -> 8 ways ( 0 to 9 excluding 1st and 2nd digit)
4th digit -> 7 ways ( 0 to 9 excluding 1st , 2nd and 3rd digit)
So total no. of ways = 9*9*8*7 = 4536 ways

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

Let's say the question would have said that the first number could as well be a zero, then the number of identification numbers would be : 10*9*8*7 ? Am i going right?

__________________
Yes, that's correct.

There are 9 choices for the first digit (1 through 9, inclusive). The second digit can be any of the 10 digits (0 through 9, inclusive) EXCEPT it can’t repeat the first digit; thus, there are 9 options for the second digit. The third digit can’t repeat either of the first two digits, so there are 8 options. Similarly, the fourth digit can’t repeat any of the first 3 digits, so there are 7 options. Thus, the total number of options is 9 x 9 x 8 x 7 = 4,536.

Answer: B

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.

9 x 9 x 8 x 7 = 4536
1 0
2 2 2
3 3 3 3
4 4 4 4
5 5 5 5
6 6 6 6
7 7 7 7
8 8 8 8
9 9 9 9

Thank you

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

Well, let’s try this...

So the code has 5 places

The first place can be selected in 9 ways as 0 cannot be included
The Second place will again take 9 ways (0 can be included but since repetition is not allowed, we can take any number excluding the first already selected and 0)
The Third place can take 8 ways and the last one will take 7 ways..
So 9*9*8*7

And the answer is B

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Identification number: _, _, _, _

Range: 0 to 9 - Total 10 numbers

The first digit cannot be zero and hence it has 9 options.

The second digit can be zero but one of the digits is fixed in the first place and hence it has 9 options.

The third digit has 8 options.

The fourth digit has 7 options.

=> 9 * 9 * 8 * 7 = 4,536

Answer B