This is just the combination formula (nCr) found in a statistics book:
27!/5!(27-5)! ! stands for factorial .
so 27! = 10888869450418352160768000000
and 5!=120
and 22!=1124000727777607680000
ans=80730
note you would hardly ever write the factorial answers out, instead just hit the n! button on your calculator to get a factorial of a number i.e to get 27! just type 27 then n!.
27
27^27
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 27 possible combinations.
There are 5C3 = 5*4/(2*1) = 10 combinations
27
31C5 which is 31!/[5!*(31-5)!] = 31*30*29*28*27/(5*4*3*2*1) = 169911
There are 5C3 = 10 combinations.
There are 5,461,512 such combinations.
You can make 5 combinations of 1 number, 10 combinations of 2 numbers, 10 combinations of 3 numbers, 5 combinations of 4 numbers, and 1 combinations of 5 number. 31 in all.
5*4*3*2*1 = 120 combinations. * * * * * No. The previous answerer has confused permutations and combinations. There are only 25 = 32 combinations including the null combinations. There is 1 combination of all 5 numbers There are 5 combinations of 4 numbers out of 5 There are 10 combinations of 3 numbers out of 5 There are 10 combinations of 2 numbers out of 5 There are 5 combinations of 1 numbers out of 5 There is 1 combination of no 5 numbers 32 in all, or 31 if you want to disallow the null combination.
27
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
Assuming there are 5 check boxes and the respondent may check any number of them, the answer is 25 = 32 combinations.
The number of combinations of r numbers out of n is nCr = n!/[r!*(n-r)!] where n! = 1*2*3...*n So here n = 30, r = 5 which gives 30*29*28*27*26/(5*4*3*2*1) = 142,506
7C3 = 7*6*5/(3*2*1) = 35 combinations.7C3 = 7*6*5/(3*2*1) = 35 combinations.7C3 = 7*6*5/(3*2*1) = 35 combinations.7C3 = 7*6*5/(3*2*1) = 35 combinations.
32 of them.