Question about Computers & Internet

Are all five numbers coming from the pool for 35 numbers?

if yes the combinations would be

35 x 34 x 33 x 32 x 31

38,955,840

if not from one pool but five different pools for 35 then it would be 35 to the 5th power

or

35 x 35 x 35 x 35 x 35

52,521,875

Posted on Oct 02, 2008

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Posted on Jan 02, 2017

I would say that the answer would be 11!/2

Feb 13, 2017 | Homework

I assume that repetition of numbers can't happen, and the order of numbers does not matter

So it is n! / (r! (n-r)!)

= 15! / (5! (15-5)!)

= 1307674368000 / (120 (3628800)

= 3003 combinations

So it is n! / (r! (n-r)!)

= 15! / (5! (15-5)!)

= 1307674368000 / (120 (3628800)

= 3003 combinations

Feb 13, 2015 | Computers & Internet

Too many to list (3,125). It would take you forever to get it unlocked. I would suggest bolt cutters and just spend the 20-30 bucks to buy a new one.

Aug 22, 2014 | Kensington Black Portable Combination Loc

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

No. There are 575,757 possible combinations. Even if the lottery was being held once an hour it would take over 65 years for all possible combinations to appear (and that assumes there are no repeats). With daily drawings, it will take over 1,500 years.

Jan 24, 2014 | Office Equipment & Supplies

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

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

If you only take one number, there are six combinations.

If you take two numbers, there are fifteen combinations.

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.

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.

If you take two numbers, there are fifteen combinations.

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.

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 | Home Security

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

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