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A company plans to assign identification numbers to its employees. Each number is to consist of four different digits from 0 to 9, inclusive, except that the first digit cannot be 0. How many different identification numbers are possible?

A)3,024
B)4,536
C)5,040
D)9,000
E)10,000

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.
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judokan
A company plans to assign identification numbers to its employees. Each number is to consist of four different digits from 0 to 9, inclusive, except that the first digit cannot be 0. How many different identification numbers are possible?

(A) 3,024
(B) 4,536
(C) 5,040
(D) 9,000
(E) 10,000

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 if 6.

Answer: B.

Hope it's clear.

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?
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Bunuel
judokan
A company plans to assign identification numbers to its employees. Each number is to consist of four different digits from 0 to 9, inclusive, except that the first digit cannot be 0. How many different identification numbers are possible?

(A) 3,024
(B) 4,536
(C) 5,040
(D) 9,000
(E) 10,000

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 if 6.

Answer: B.

Hope it's clear.

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

Only answer with units digit 6 is B.

Hope this helps
Cheers
J
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judokan
A company plans to assign identification numbers to its employees. Each number is to consist of four different digits from 0 to 9, inclusive, except that the first digit cannot be 0. How many different identification numbers are possible?

(A) 3,024
(B) 4,536
(C) 5,040
(D) 9,000
(E) 10,000

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
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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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judokan
A company plans to assign identification numbers to its employees. Each number is to consist of four different digits from 0 to 9, inclusive, except that the first digit cannot be 0. How many different identification numbers are possible?

(A) 3,024
(B) 4,536
(C) 5,040
(D) 9,000
(E) 10,000

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

So different identification numbers possible : 4536
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judokan
A company plans to assign identification numbers to its employees. Each number is to consist of four different digits from 0 to 9, inclusive, except that the first digit cannot be 0. How many different identification numbers are possible?

(A) 3,024
(B) 4,536
(C) 5,040
(D) 9,000
(E) 10,000

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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Bunuel
A company plans to assign identification numbers to its employees. Each number is to consist of four different digits from 0 to 9, inclusive, except that the first digit cannot be 0. How many different identification numbers are possible?

(A) 3,024
(B) 4,536
(C) 5,040
(D) 9,000
(E) 10,000

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 if 6.

Answer: B.

Hope it's clear.

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?
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Bunuel
A company plans to assign identification numbers to its employees. Each number is to consist of four different digits from 0 to 9, inclusive, except that the first digit cannot be 0. How many different identification numbers are possible?

(A) 3,024
(B) 4,536
(C) 5,040
(D) 9,000
(E) 10,000

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 if 6.

Answer: B.

Hope it's clear.

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?
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Yes, that's correct.
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judokan
A company plans to assign identification numbers to its employees. Each number is to consist of four different digits from 0 to 9, inclusive, except that the first digit cannot be 0. How many different identification numbers are possible?

(A) 3,024
(B) 4,536
(C) 5,040
(D) 9,000
(E) 10,000

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
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judokan
A company plans to assign identification numbers to its employees. Each number is to consist of four different digits from 0 to 9, inclusive, except that the first digit cannot be 0. How many different identification numbers are possible?

(A) 3,024
(B) 4,536
(C) 5,040
(D) 9,000
(E) 10,000

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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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
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judokan
A company plans to assign identification numbers to its employees. Each number is to consist of four different digits from 0 to 9, inclusive, except that the first digit cannot be 0. How many different identification numbers are possible?

(A) 3,024
(B) 4,536
(C) 5,040
(D) 9,000
(E) 10,000

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

Posted from my mobile device
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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
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