Help! How Many Variations of a 4 Digit Number Exist?

[NellnBear]

Hi, I'm Jenell... I am just joining so maybe someone can help me out with something I'm curious about. :)

Okay... so My husband asked me to figure out how many variations of a 4 digit number (like a pin number) you could come up with... and without writing it all down I was hoping to figure it out... He's trying to challenge me since I just did really good in my math class this last quarter. lol

I came up with the following... am I completley wrong? Anyone know the answer?

take 1000, 1100, 1200, 1300, 1400, 1500, 1600, 1700, 1800, 1900 & 99 an add them all up... because that would cover all of the variations of 1000 that you could use... and then times that by 10 because there are 10 numbers on a key pad... and then you'd get 145,990 variations...

Anyone know? Help!

 

[NellnBear]

Oops I didn't mean to post in this folder. :)

 

[shmoe]

Looking at your answer, I really don't think I understand your question. By 'variation' do you mean all possible 4 digit pins? Or do you mean some sort of rearrangement of a given pin number? Something else?

If you asked the same question for 1 digit, what would your answer be? 2 digits? 3 digits? If you can answer these it might clarify (it might help you come up with a solution for the 4 digit case as well!)

 

[Data]

Do you mean how many four digit numbers are there? In that case, it's just 10^4 = 10000 (choose 4 digits, with 10 choices for each).

On the other hand, if each digit has to be unique (ie. you can't have two of the same digit), then your answer requires factorials.

For any nonnegative integer (natural number, ie. 0, 1, 2, 3, ...) n, define the factorial of n, denoted by n!, by

n! = n(n-1)(n-2). \ . \ . (3)(2)(1)

if n>0 and define 0! = 1.

Now note that if we need to pick 4 unique digits, then there are 10 choices for the first digit, 9 choices for the second digit, 8 for the third, and 7 for the fourth. Thus there are

(10)(9)(8)(7) = \frac{10!}{6!} = 5040

ways to choose the number.

Another possibility is that you need to know the number of ways in which you can order a given 4 digits. This is a similar problem to the last one: there are 4 choices for the first digit, 3 for the second, 2 for the third, and only 1 for the last, so in total there are 4! = 24 ways to order them.

 

[NellnBear]

Hmmm now I'm really not sure! Basically I guess how I could put it would be how many different 4 digit numbers could you come up with... But 10000 doesn't seem right... that seems way to easy. Ha!

 

[Data]

Well, in that case, you have precisely 10 choices for each digit, and you're choosing 4 of them.

I have 10 choices for the first digit. Once I choose the first digit, I have 10 choices for the second digit. But there were 10 ways to pick the first one, so in total I have (10)(10)=100 ways to choose the two digits. Induct as necessary.

 

[matt grime]

Erm, or you could just have any of the numbers 0000 to 9999 inclusive, and there are how many of those?

 

[Data]

Hahah. I'm brainwashed by probability classes.

 

[xanthym]

Or perhaps neither Order nor Uniqueness of digits matter. Such would be the case, for example, if the PINs "1777", "7177", "7717", & "7771" were all acceptable but were all considered to be the same selection (all counting like 1 PIN). The number of such selections would then be the number of combinations of 4 chosen from 10 distinct items with repetition:
{Combinations of 4 From 10 w/Rep} = C{(10 + 4 - 1), 4} =
= (10 + 4 - 1)!/{(4!)*(10 + 4 - 1 - 4)!} =
= (13)!/{(4!)*(9!)} =
= 715


~~

 

[robert Ihnot]

I was thinking of the question like a pick-4 in the lottery daily drawing. A person there, to conserve betting capital, picks his four digits, say: 1,2,3,4, and looks at all the different ways that can occur, which is 4!=24. Now if two digits are the same, then the answer is: 4!/2! = 12, and so forth. Obviously if all the digits are the same the answer is 4!/4! =1, which is the same as "playing it straight."

 

[slackr007]

I'm pretty sure matt grime's post sums up the problem. you've got numbers 0000 through 9999 inclusive, so that's 9999+1=10000. That's what you were asking, right? That's how many 4 digit numbers there are.

 

[DNGO]

Do you mean how many four digit numbers are there? In that case, it's just 10^4 = 10000 (choose 4 digits, with 10 choices for each).

On the other hand, if each digit has to be unique (ie. you can't have two of the same digit), then your answer requires factorials.

For any nonnegative integer (natural number, ie. 0, 1, 2, 3, ...) n, define the factorial of n, denoted by n!, by

n! = n(n-1)(n-2). \ . \ . (3)(2)(1)

if n>0 and define 0! = 1.

Now note that if we need to pick 4 unique digits, then there are 10 choices for the first digit, 9 choices for the second digit, 8 for the third, and 7 for the fourth. Thus there are

(10)(9)(8)(7) = \frac{10!}{6!} = 5040

ways to choose the number.

Another possibility is that you need to know the number of ways in which you can order a given 4 digits. This is a similar problem to the last one: there are 4 choices for the first digit, 3 for the second, 2 for the third, and only 1 for the last, so in total there are 4! = 24 ways to order them.

Becareful with your logic on how many ways you can order a given 4 digits. For example, consider the numbers 1111, 9999, 2222, 0000 which are part of the answer to the original problem of "How many 4-digit numbers there are". How many ways can you order any of the numbers just given above? The answer is 1.