How many different combinations are there for the numbers 1, 2, 3, 3, 4, 5 and 6?

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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.

Posted on Apr 30, 2013

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Posted on Jan 02, 2017

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

Feb 13, 2017 | Homework

Each Brinks 5059 safe has a different default combination, so to get the original combination you must either contact Honeywell (the parent company for Brinks safes) at 1-800-223-8566.

You need to send them a notarized letter showing safe ownership in order to get the master combination. When you call Honeywell, you will need to provide your safe's model number, the serial number (found on the right of the safe, near the model number), and your key number.

You need to send them a notarized letter showing safe ownership in order to get the master combination. When you call Honeywell, you will need to provide your safe's model number, the serial number (found on the right of the safe, near the model number), and your key number.

Jan 04, 2017 | Home Security

I assume you mean different decimal numbers, using those four digits, right? If so, the answer can be reached two ways.

One way is to actually build up the model by changing one digit at a time, then moving a decimal place. Thus: 1108, 1101, 1100 then 1118, 1111, 1110 then 1188, 1181, 1180 and so on. You will count 81 combinations.

Since those three different digits can be placed in four decimal places, the answer in another way is: 3*3*3*3=81.

One way is to actually build up the model by changing one digit at a time, then moving a decimal place. Thus: 1108, 1101, 1100 then 1118, 1111, 1110 then 1188, 1181, 1180 and so on. You will count 81 combinations.

Since those three different digits can be placed in four decimal places, the answer in another way is: 3*3*3*3=81.

Sep 27, 2014 | Office Equipment & Supplies

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

1,048,575.

There are 20 different 1-number combinations, 190 different 2-number combinations, 1140 three-number combinations, and so on.

There are 20 different 1-number combinations, 190 different 2-number combinations, 1140 three-number combinations, and so on.

Mar 17, 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

Buy the identical lock if possiable and return as defective and they will reset. Otherwise odds or 1 in 999,000 differant combos. This is airport baggage security lock.

Please rate. Thanks

Please rate. Thanks

Feb 21, 2009 | Hardware & Accessories

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

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

Oct 02, 2008 | Computers & Internet

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