Factorial Equation set up - trying to set up an equation to determine
how many different integers can be formed by using a five digit
number, with one recurring digit (such as 95254),with the largest
integer being no more than five digits long.

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Request for Question Clarification by endo-ga on 01 Dec 2003 22:23 PST

Hi,

To start off for the digit choice, assuming you mean 1 and only 1
repeated digit and that your integer can start with 0 you've got
10 for the first
9 for the second
8 for the third
7 for the fourth
and only 1 for the fifth, the one you repeat.

Now you've just need to find all the ways these can be arranged to
form different integers.

Thanks.
endo

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Clarification of Question by dclnop-ga on 02 Dec 2003 06:36 PST

I (think) I know that to find the number of five digit combinations of
a given five digit number, the equation is 5!, assuming there are no
repetitive digits.  My conundrum is: 1.  how to find the number of
combinations given that one of the digits in the original number
repeats, and 2.  the number of combinations of 1,2,3 and 4 digit
numbers that can be formed using the original five digit number with
one repeating digit.

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Request for Question Clarification by endo-ga on 02 Dec 2003 08:30 PST

Hi,

I think you just need to multiply the result by 4. The only digit that
needs to be placed is the digit that is repeated. All other
possibilities are already included.

Thanks.
endo

Hello and thanks for your interesting question.

The answer depends on whether a 5-digit number can start with '0'

If it can, then here's the pattern
 0  0  [9] [8] [7]   first is 0, second is 0, 9x8x7=504 ways to fill the rest
 0 [9]  0  [8] [7]   another 504 ways
 0 [9] [8]  0  [7]          "
 0 [9] [8] [7]  0           "
[9] 0   0  [8] [7]          "
[9] 0  [8]  0  [7]          " 
[9] 0  [8] [7]  0           "
[9][8]  0   0  [7]          " 
[9][8]  0  [7]  0           "
[9][8][7]   0   0           "
  So the total of 5-digit #s with a double '0' and no other double is 5,040

Then repeat that exact pattern using '1' where the '0' used to be
and likewise for '2' through '9'

Grand total = 10 * 5,040 = 50,400

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When a leading '0' is not allowed, the '0' group is only six rows high
instead of ten.
---------

No google search terms this time--just used my noodle.

Cheers!
Richard-ga

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Clarification of Answer by richard-ga on 03 Dec 2003 13:45 PST

dclnop-ga:

You'll want to consider racecar-ga's comment, below, which has an
entirely different reading of your question.

-R

It appears to me that dclnop-ga wants the number of distict ways that
the digits in a given 5 digit number (with one repeating digit) can be
arranged, and not the number of distinct 5 digit numbers with one
repeating digit that can be formed using all 10 digits.

If we assume that the given 5 digit number has no zeros in it (say
it's 11234), then there are:

2 digit numbers: 4*3 + 1 = 13  
  (4 choices for the first digit, 3 for the second, plus the number 11)

3 digit numbers: 4*3*2 + 3*3 = 33
  (4 choices for the first digit, 3 for the second, and 2 for the
third, plus the numbers like 112, for which there are 3 choices for
the third digit, and 3 choices for where to place the third digit)

4 digit numbers: 4*3*2*1 + (4 choose 2)*3*2 = 60
  ( (4 choose 2)=6 is the number of ways to place the two 1's, and
then there are 3 choices for the third digit and 2 for the fourth)

5 digit numbers: (5 choose 2)*3*2*1 = 60
  ( (5 choose 2) = 10 is the number of ways to place the 1's...)

So the answer is 166.

If one of the digits is a zero (like in 11230), then

3 of the 2 digit numbers would start with a 0,
7 of the 3 digit numbers   "    "    "    "
12 of the 4 digit numbers  "    "    "    "
12 of the 5 digit numbers  "    "    "    "

So in this case, the answer is 132

If the repeated digit is a zero (like in 12300), then

4 of the 2 digit
12 of the 3 digit
24 of the 4 digit
24 of the 5 digit numbers would start with a zero.

So in this case, the answer is 102