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A company plans to assign identification numbers

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nanu.nantaki
 Post subject: A company plans to assign identification numbers
 Post Posted: Wed Sep 09, 2009 1:02 am 
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Students


Posts: 7
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 answer is B, I'm not sure, how. Please help !! How can I find answer using MGMAT method. Thanks.
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ankitmisri
 Post subject: Re: A company plans to assign identification numbers
 Post Posted: Wed Sep 09, 2009 7:17 am 
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Posts: 4
1. Each number is to consist of four different digits from 0 to 9
Hence no number can repeated
2. the first digit cannot be 0

hence you can only choose 1 of the 9 numbers(1-9) for the 1st place i.e. is you have 9 options for 1st digit

For the 2nd digit again you can again choose 1 out 9 numbers (now u have 0 available) in 9c1 ways =9

For the 3rd digit only 8 numbers are available coz of "1."...u can choose that in 8c1 ways = 8

For the 4th place only 7 numbers are available coz of "1."...u can choose that in 7c1 ways = 7

Since all four selected individually, you have to multiply them 9*9*8*7 = 4536
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andersonbae
 Post subject: Re: A company plans to assign identification numbers
 Post Posted: Wed Sep 09, 2009 7:29 am 
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Posts: 3
Here is an easy solution , if you don't know how to solve this problem.
Don't be embarrassed.

There are three numbers.
0, 1, 2

We can make the identification numbers which does not start "0" as the following

102
120
201
210

The total number of the identification numbers we can make is 4 with 3 different numbers.

We can apply a "SLOT" solution here.

--- --- ---

(3-1) => First Slot : because "O" should not start as a first number
2 => Second Slot
1 => Third Slot

2*2*1 = 4


We apply this way to the original problem.
(10-1) - because "0" should not start as a first number.

(10-1) * 9 * 8 * 7 = 4536

Does it make sense? If I am wrong, please let me know.
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aaron.anglada
 Post subject: Re: A company plans to assign identification numbers
 Post Posted: Mon Sep 14, 2009 10:18 am 
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Course Students


Posts: 2
I like this method:

First, find out the number of ways you can create the ID numbers WITHOUT any restrictions, then subtract the number of ways we create ID numbers WITH the restriction.

That will give us the number of ways that we can create only valid ID numbers

0->9 inclusive is really 10 digits

The number of ways we can arrange 10 digits in four "slots" is 10 * 9 * 8 * 7 = 5040

So there are 5040 unrestricted numbers

We can find the number of restricted ID numbers by assuming the first digit is already 0. That leaves nine digits left for three "slots":

9 * 8 * 7 = 504

unrestricted ID numbers - restricted ID numbers = Possible ID numbers

5040 - 504 = 4536
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RonPurewal
 Post subject: Re: A company plans to assign identification numbers
 Post Posted: Sat Sep 26, 2009 1:57 am 
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ManhattanGMAT Staff


Posts: 7146
ankitmisri wrote:
1. Each number is to consist of four different digits from 0 to 9
Hence no number can repeated
2. the first digit cannot be 0

hence you can only choose 1 of the 9 numbers(1-9) for the 1st place i.e. is you have 9 options for 1st digit

For the 2nd digit again you can again choose 1 out 9 numbers (now u have 0 available) in 9c1 ways =9

For the 3rd digit only 8 numbers are available coz of "1."...u can choose that in 8c1 ways = 8

For the 4th place only 7 numbers are available coz of "1."...u can choose that in 7c1 ways = 7

Since all four selected individually, you have to multiply them 9*9*8*7 = 4536


yep. this is the most direct method, and thus the best.

if you have a problem in which ORDER MATTERS, then you should be able to solve the problem by multiplying numbers together.
you shouldn't have to bother with factorials, etc., unless you have a problem on which order DOESN'T matter.
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