Question about Microsoft Computers & Internet

How many combinations of 6 numbers are there in the 20 numbers

as set out here . 2.4.6.12.18.20.24.26.28.30

5.7.9.15.17.21.23.25.29.31

Posted on Nov 10, 2013

Dear Sir/madam,

Read out the formula with examples of the Ms-word Help Menu. PERMUT, COMBIN. It will all teach you......

thanks

good luck

Posted on Aug 14, 2008

U have to go with a professional

Posted on Aug 14, 2008

Hi,

Have you tried "PERMUT" using MS Excel?

Hope this be of initial help/idea. Pls post back how things turned up or should you need additional information.

Good luck and kind regards. Thank you for using FixYa.

Posted on Aug 13, 2008

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Posted on Jan 02, 2017

Do you want combinations or permutations? They're different things. A combination of n items taken m at a time, sometimes written "nCm", selects m items from a group of n without regard to order. A permutation of n items taken m at a time, sometimes written "nPm", selects m items from a group of m with regard to order.

Given six numbers, there are six combinations of one number, 15 combinations of two numbers, 20 combinations of three numbers, 15 combinations of four numbers, six combinations of five numbers, and one combination of six numbers.

Given six numbers, there are six permutations of one number, 30 permutations of two numbers, 120 permutations of three numbers, 360 permutations of four numbers, 720 permutations of five numbers, and 720 permutations of six numbers.

Given six numbers, there are six combinations of one number, 15 combinations of two numbers, 20 combinations of three numbers, 15 combinations of four numbers, six combinations of five numbers, and one combination of six numbers.

Given six numbers, there are six permutations of one number, 30 permutations of two numbers, 120 permutations of three numbers, 360 permutations of four numbers, 720 permutations of five numbers, and 720 permutations of six numbers.

Oct 22, 2014 | Office Equipment & Supplies

There is only one combination using all five digits, namely 1, 2, 3, 4, 7 in some order.

If the order matters then there are 120 different permutations. I won't list them all here, but they start with 1, 2, 3, 4, 7 and go through 7, 4, 3, 2, 1.

If the order matters then there are 120 different permutations. I won't list them all here, but they start with 1, 2, 3, 4, 7 and go through 7, 4, 3, 2, 1.

Aug 21, 2014 | Computers & Internet

One-number combinations: { 1 2 6 9 }

Two-number combinations: { 12 16 19 26 29 69 }

Three-number combinations: { 126 129 169 269 }

Four-number combinations { 1269 }

Or did you want permutations?

Two-number combinations: { 12 16 19 26 29 69 }

Three-number combinations: { 126 129 169 269 }

Four-number combinations { 1269 }

Or did you want permutations?

Jul 07, 2014 | Computers & Internet

There is only one six-digit combination of those six digits, namely all six digits: 1, 4, 5, 7, 8, 9

There are 720 permutations if the order matters. I won't list them all here.

There are 720 permutations if the order matters. I won't list them all here.

Apr 09, 2014 | Computers & Internet

There's only one six-digit combination possible, namely the digits 1, 2, 3, 4, 5, and 6.

If the order matters then there are 720 possible permutation. I'm not going to list them all. Start with 123456, then 123465, 123546, and so on to 654312 and 654321.

If the order matters then there are 720 possible permutation. I'm not going to list them all. Start with 123456, then 123465, 123546, and so on to 654312 and 654321.

Mar 25, 2014 | Computers & Internet

If the order of the tree digits matter:

9!/6! = 504

If the order doesn't matter:

9!/(6!*3!) = 84

For more info, research combinations and permutations

9!/6! = 504

If the order doesn't matter:

9!/(6!*3!) = 84

For more info, research combinations and permutations

Feb 03, 2014 | Mathsoft Computers & Internet

I need all possible combinations of 9 numbers in 7-6-5 digits groups with list of combinations chart?

Jan 06, 2014 | Computers & Internet

That depends on how many of those six numbers you take.

If you only take one number, there are six combinations: {1}, {2}, {3}, {4}, {5}, and {6}.

If you take two numbers, there are fifteen combinations: {1,2}, {1,3}, {1,4}, {1,5}, {1,6}, {2,3}, {2,4}, {2,5}, {2,6}, {3,4}, {3,5}, {3,6}, {4,5}, {4,6}, and {5,6}.

If you take three numbers, there are twenty combinations.

If you take four numbers, there are fifteen combinations.

If you take five numbers, there are six combinations.

If you take all six numbers, there is only one combination: {1,2,3,4,5,6}.

In general, if you take 'm' objects out of a set of 'n' objects, the number of combinations is given by n!/[(m!)(n-m)!] where '!' is the factorial operator.

If you only take one number, there are six combinations: {1}, {2}, {3}, {4}, {5}, and {6}.

If you take two numbers, there are fifteen combinations: {1,2}, {1,3}, {1,4}, {1,5}, {1,6}, {2,3}, {2,4}, {2,5}, {2,6}, {3,4}, {3,5}, {3,6}, {4,5}, {4,6}, and {5,6}.

If you take three numbers, there are twenty combinations.

If you take four numbers, there are fifteen combinations.

If you take five numbers, there are six combinations.

If you take all six numbers, there is only one combination: {1,2,3,4,5,6}.

In general, if you take 'm' objects out of a set of 'n' objects, the number of combinations is given by n!/[(m!)(n-m)!] where '!' is the factorial operator.

Apr 30, 2013 | Mathsoft Computers & Internet

Hi there!The simple formula for permutations is n^r - in this case, it would be 10^6 which is 1,000,000. This is the answer if repetitions are allowed.
If repetitions are NOT allowed, then the formula changes. It would then be 10 * 9 * 8 * 7 * 6 * 5 = 151,200, which is also 10!/4!
Hope that helped!
Alex

May 31, 2011 | Computers & Internet

I assume you are looking for the number of combinations of 1 - 45 where the first number is one of the 45, the second number is one of the remaining 44, etc. The answer is 45 factorial (i.e x=45*44*43*...*2*1). The Excel formula Is =FACT(45).

Jul 09, 2009 | Microsoft Excel for PC

May 25, 2017 | Microsoft Computers & Internet

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Usually answered in minutes!

Thanks for your help but I know how to calculate how many permutations I will get what I need is to list each one of them out in 6 columns on an excel spreadsheet so I need formula to post that I can fill down in each column.

eg 111111, 111112, 111113, 111114 ... 223456 etc and so on there should be 9x9x9x9x9x9=531,441 permutations in my final list with none repeating.

I am using Microsoft excel version 2007. The CHAR I refer to is a formula option like ROUND or PERMUT. Sorry for the confusion.

I have been e-mailed to tell me that there is a new comment on this problem but there is nothing new here!!

I have not really had a response to this problem yet. Can anyone help as I need to get this list pronto!!

Thanks to all who have answered but you have all told me how to calculate how many permutations which I already know. I need formula to put at the top of each column in cells a1 to f1 which I can then fill down to give me a long column of all the combinations

eg

111111

111112

111113

111114

I have the formula for picking three alphabetical characters from ABCDEFGHIJK.

In cell A1 type =CHAR(65+TRUNC((ROW()-1)/144))

In cell B1 type =CHAR(65+MOD(TRUNC((ROW()-1)/12),12))

In cell C1 type =CHAR(65+MOD(ROW()-1,12))

This does work but I do not know how to expand it to 6 columns with 9 options. I know that CHAR 49 is the ASCII ref for number 1 and 50 is for number 2 etc so I can substitute that and I assume that the 12 will be replaced with 9 but apart from that I do not understand the logic of the formula.

I hope this helps you find me a solution.

Sue

Sorry I am new to the site. If an expert just sends a comment that says I know tha answer how do I look at it without knowing what it is?

Solved problem myself thankyou. No need to further respond.

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