A company plans to assign identification numbers

[nanu.nantaki]

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.

[ankitmisri]

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

[andersonbae]

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.

[aaron.anglada]

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

[RonPurewal]

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.

[ajaym8]

Hi Ron,
What is wrong if I do it as indicated below ?

Asking because this method came to my mind first in the mock and I could not get to the correct answer.

I have 4 places to fill.
_ _ _ _

I start with the last digit(one's place), I have 10 numbers to chose from. Similarly for 3rd digit I have 9 numbers to chose from, 8 numbers for the 2nd place.
Now for the first place I cannot include 0, so I have 6 numbers to chose from(7 numbers if 0 were allowed) for the first place (the thousand's place).

I get 6*8*9*10. which gives 4320.

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.

[RonPurewal]

the problem with your approach is that there might be 6 possibilities for the thousands place, but there also might be 7 possibilities.

specifically,
• if one of the last three digits -- the ones you've already chosen -- is 0, then there are still 7 digits available for the thousands place (first slot).
• if none of the last three digits is 0, then, there are only 6 digits available for the first slot.

therefore you can't multiply the possibilities in this order -- unless you want to break them into 2 separate cases. but that would be super annoying, because then you'd have to think about exactly how many different ways you could pick the last three digits with 0 among them, and how many different ways you could pick the last three digits without 0.

[RonPurewal]

more generally, you should ALWAYS pick the items that have RESTRICTIONS ••FIRST•• in problems like this.
otherwise, you get the problem you have here -- if you leave the restricted places for last, then you can't calculate the numbers of possibilities for them (because that depends on the previous choices).