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

  • Thread starter NellnBear
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In summary: But using the logic ((10)(9)(8)(7)) = 5040, you get 5040 ways.In summary, there are 10000 possible 4-digit numbers (0000 to 9999), but if each digit has to be unique, then there are 5040 possible combinations. If order does not matter, then there are 715 possible combinations. The original question was not clear, so it is difficult to determine which answer is the correct one.
  • #1
NellnBear
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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! :confused:
 
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  • #2
Oops I didn't mean to post in this folder. :)
 
  • #3
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!)
 
  • #4
Do you mean how many four digit numbers are there? In that case, it's just [itex]10^4 = 10000[/itex] (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 [itex]n![/itex], by

[tex] n! = n(n-1)(n-2). \ . \ . (3)(2)(1)[/tex]

if [itex] n>0[/itex] and define [itex]0! = 1[/itex].

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

[tex](10)(9)(8)(7) = \frac{10!}{6!} = 5040[/tex]

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 [itex]4! = 24[/itex] ways to order them.
 
  • #5
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!
 
  • #6
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.
 
  • #7
Erm, or you could just have any of the numbers 0000 to 9999 inclusive, and there are how many of those?
 
  • #8
Hahah. I'm brainwashed by probability classes.
 
  • #9
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


~~
 
  • #10
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."
 
  • #11
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.
 
  • #12
Data said:
Do you mean how many four digit numbers are there? In that case, it's just [itex]10^4 = 10000[/itex] (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 [itex]n![/itex], by

[tex] n! = n(n-1)(n-2). \ . \ . (3)(2)(1)[/tex]

if [itex] n>0[/itex] and define [itex]0! = 1[/itex].

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

[tex](10)(9)(8)(7) = \frac{10!}{6!} = 5040[/tex]

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 [itex]4! = 24[/itex] 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.
 

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

How do you calculate the number of variations of a 4 digit number?

The number of variations of a 4 digit number can be calculated by using the formula n^r, where n is the number of options for each digit and r is the number of digits. In this case, n would be 10 (since there are 10 possible digits - 0 to 9) and r would be 4. So the total number of variations would be 10^4 = 10,000.

Is there a difference between variations and combinations in a 4 digit number?

Yes, there is a difference between variations and combinations. Variations refer to the number of different ways a set of elements can be arranged, while combinations refer to the number of different ways a subset of elements can be selected from a larger set. In the case of a 4 digit number, variations would include all possible combinations of the 4 digits (such as 1234, 2341, 3412, etc.), while combinations would only include unique sets of 4 digits (such as 1234, 5678, etc.).

Can you give an example of a 4 digit number with the maximum number of variations?

Yes, the maximum number of variations for a 4 digit number would be 10^4 = 10,000. An example of such a number would be 9999, as all 4 digits are different and there are no repeating digits.

How many variations of a 4 digit number are possible if repetition is not allowed?

If repetition is not allowed, the number of variations of a 4 digit number would be nPr, where n is the number of options and r is the number of digits. In this case, n would be 10 and r would be 4. So the total number of variations would be 10P4 = 10 * 9 * 8 * 7 = 5040.

Can the number of variations of a 4 digit number be greater than 10,000?

No, the number of variations of a 4 digit number cannot be greater than 10,000. This is because there are only 10 possible digits (0 to 9) and we are dealing with a 4 digit number, so the maximum number of variations would be 10^4 = 10,000.

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