You question is quite vague. Do you mean that you have 27 numbers, and how many possible combinations are available if you can only take seven of those 27 at a time? If so 4,475,671,200 (four billion, four hundred and seventy five million, six hundred and seventy one thousand and two hundred)
There are 167960 9 digits combinations between numbers 1 and 20.
there are 13,983,816 combinations.
There are 1140 five digit combinations between numbers 1 and 20.
8
There is 1 combinations with 7 numbers in them,7 combinations with 1 or 6 numbers in them,21 combinations with 2 or 5 numbers in them,35 combinations with 3 or 4 numbers in them.Notionally, there is also one combination with no numbers in it.In all then, there are 128 ( = 27) combinations.There is 1 combinations with 7 numbers in them,7 combinations with 1 or 6 numbers in them,21 combinations with 2 or 5 numbers in them,35 combinations with 3 or 4 numbers in them.Notionally, there is also one combination with no numbers in it.In all then, there are 128 ( = 27) combinations.There is 1 combinations with 7 numbers in them,7 combinations with 1 or 6 numbers in them,21 combinations with 2 or 5 numbers in them,35 combinations with 3 or 4 numbers in them.Notionally, there is also one combination with no numbers in it.In all then, there are 128 ( = 27) combinations.There is 1 combinations with 7 numbers in them,7 combinations with 1 or 6 numbers in them,21 combinations with 2 or 5 numbers in them,35 combinations with 3 or 4 numbers in them.Notionally, there is also one combination with no numbers in it.In all then, there are 128 ( = 27) combinations.
There are 59C5 = 59*58*57*56*55/(5*4*3*2*1) = 5,006,386 combinations.
I need to have listed all the 4 number combinations between 1 and 9
There are 6C3 = 20 such combinations.
There are 5,461,512 such combinations.
There are millions of possible combinations.
157.
The number of combinations is 32!/[6!*(32-6)!] where n! represents 1*2*3*...*n The answer is 32*31*30*29*28*27/(6*5*4*3*2*1) = 906192