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.
To find the combinations of 5 bills and 2 bills that sum to 27, we can denote the number of 5-dollar bills as (x) and the number of 2-dollar bills as (y). The equation we need to solve is (5x + 2y = 27) with the constraints (x \geq 0) and (y \geq 0). By testing non-negative integer values for (x), we find that possible combinations are ( (5, 1), (4, 3), (3, 5), (2, 7), (1, 9), (0, 13) ), resulting in a total of 6 combinations.
There are 27 possible combinations.
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.
31C5 which is 31!/[5!*(31-5)!] = 31*30*29*28*27/(5*4*3*2*1) = 169911
There are 27 possible combinations.
There are 5C3 = 5*4/(2*1) = 10 combinations
There are 5,461,512 such combinations.
There are 5C3 = 10 combinations.
To determine the number of different combinations of coins that make up 27 cents, we can use pennies (1 cent), nickels (5 cents), dimes (10 cents), and quarters (25 cents). A systematic approach or generating functions can be employed, but an approximate method indicates that there are 13 combinations using these coins. The combinations include various configurations of each coin type to total 27 cents.
There are 2*2*5 = 20 combinations.
27
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.