CA United States Member #2987 December 10, 2003 832 Posts Offline

Posted: June 30, 2009, 11:35 am - IP Logged

I just did an Excel chart similar to this but for Pick 3. Pick 4 would be 10 times as large, of course. I presently have no plans to work on something like this, as 10,000 cells would have to be individually formatted. But:

There are 625 4-digit combinations that can be formed using 4 digits from the first 5-digit group. This includes doubles, triples and quads. There are 2,500 4-digit combinations that can be formed using 3 digits from the first 5-digit group and 1 digit from the second 5-digit group. This includes doubles and triples. There are 3,750 4-digit combinations that can be formed using 2 digits from the first 5-digit group and 2 digits from the second 5-digit group. This includes doubles and combinations where there are two sets of doubles. There are 2,500 4-digit combinations that can be formed using 1 digit from the first 5-digit group and 3 digits from the second 5-digit group. This includes doubles and triples. There are 625 4-digit combinations that can be formed using 4 digits from the second 5-digit group. This includes doubles, triples and quads.

The Excel output would have to be text to allow the first digit in an individual cell to be a zero. The finished chart would have 400 rows and 25 columns, 200 rows and 50 columns, or 100 rows and 100 columns, depending on preference.

There are several ways to display the results; my personal preference is to use a formula that can be copied, then easily edited to cover all the possibilities.

Assuming the initial 5-digit group is posted in Cell A1, the first cell in which output is desired would have the formula =MID(A1,1,1)&MID(A1,1,1)&MID(A1,1,1)&MID(A1,1,1). This formula yields a quad consisting of the first number in the first 5-digit group. Note that MID is used for all output, not LEFT or RIGHT - this allows for easier editing of the copied formula to other cells. If I were continuing, the cell immediately to the right of the first output cell would have the formula =MID(A1,1,1)&MID(A1,1,1)&MID(A1,1,1)&MID(A1,2,1). This would yield a triple, consisting of the first number in the first 5-digit group three times and the second number in the first 5-digit group once.

Hope you get the idea.

gl

j

Blessed Saint Leibowitz, keep 'em dreamin' down there.....

Next week's convention for Psychics and Prognosticators has been cancelled due to unforeseen circumstances.

Kentucky United States Member #32652 February 14, 2006 7325 Posts Offline

Posted: July 1, 2009, 9:43 am - IP Logged

Quote: Originally posted by wellcrafted on June 30, 2009

Hi,

How can I figure out how many 4 digit combinations can be made from 5 digits.

Then, how do I actually GET these combinations? ( I want to put them through a filter, of course.)

Any help is appreciated.

I believe it's called factoring when any of 5 digits can appear in any of the four positions. To get the total number of combinations you multiple 5 (first position) times 5 (second) times 5 (third) times 5 ( fourth) equals 625. If you're using the digits 1-2-3-4-5 it would look like (1-2-3-4-5) with (1-2-3-4-5) with (1-2-3-4-5) with (1-2-3-4-5).

You'll probably need a program that generates all 625 combinations and allows you to apply filters.

NY United States Member #23835 October 16, 2005 3475 Posts Offline

Posted: July 21, 2009, 3:42 pm - IP Logged

Perhaps you're thinking of factorials (n!), which are a number multiplied by each whole number that is less than that number and greater than zero. For 5 it's 5*4*3*2*1, which is 120. There are 120 permutations for 5 numbers if you use all 5 of them. In a permutation the order of the numbers matters. For combinations, OTOH, the order doesn't matter. 1,2,3,4,5 and 5,4,3,2,1 are different permutations, but represent irrelevant variations of the same combination. The formulas for permutations and combinations start with the factorial of the full set of numbers, which is then reduced based on how many of the numbers are being used, and whether you want the number of permutations or combinations.

For 4 of 5 it's 5! / (4! * (5-4)! ), which simplifies to 5! / ( 4! * 1 ) --> 5! / 4! = 5.

The list above omitted the 5th combination of ABDE.

mid-Ohio United States Member #9 March 24, 2001 19831 Posts Offline

Posted: July 21, 2009, 7:49 pm - IP Logged

Quote: Originally posted by wellcrafted on June 30, 2009

Hi,

How can I figure out how many 4 digit combinations can be made from 5 digits.

Then, how do I actually GET these combinations? ( I want to put them through a filter, of course.)

Any help is appreciated.

You can down load a free program from LottoLogix called "Maybelle QuickWheel" that will calculate that it's 625 combinations and print them out for you.

* you don't need to buy more tickets, just buy a winning ticket *

California United States Member #48843 January 11, 2007 27 Posts Offline

Posted: July 23, 2009, 12:32 am - IP Logged

If you had a database program such as MS Access, you could create a table named tblRnd1 with a single field named Num of data type number (though data type of text works too). Then populate your table with the five digits.

This query right below outputs the 625 records / combos (assuming you named the table as listed above and populated it):