3!(factorial) or six
There are infinitely many possible answers. Here is one: 62 + 1
Assuming 9 numbers chosen from 56, with no repetition allowed, there are 7575968400 possible combinations.
With two bits, there are (2^2) possible combinations, which equals 4. The combinations are: 00, 01, 10, and 11. Each bit can be either 0 or 1, leading to these four distinct configurations.
At least 11 of them.
The total number of 1-bit combinations is 2. This is because a single bit can have two possible values: 0 or 1. Therefore, the combinations are {0, 1}.
There are millions of possible combinations.
There are eight possible combinations.
Just 1.
If you have six different milkshake flavors, the number of combinations you can make depends on whether you can use one flavor, multiple flavors together, and if the order of flavors matters. Assuming you can choose any combination of flavors (including using none or all), the total number of combinations is (2^6 - 1 = 63), where (2^6) represents all possible subsets of the flavors, and we subtract 1 to exclude the empty set (no milkshake). Therefore, you can create 63 different combinations of milkshakes.
There are infinitely many possible answers. Here is one: 62 + 1
Assuming 9 numbers chosen from 56, with no repetition allowed, there are 7575968400 possible combinations.
There are 1285C4 = 1285*1284*1283*1282/(4*3*2*1) = 113076300485 combinations.
We select 1 shirt out of 6 shirts in (6 choose 1) ways or 6 ways. Then, we select 1 out of 3 pairs of shorts in (3 choose 1) or 3 ways. Therefore, the possible combinations of a shirt and a pair of shorts is 6 * 3 = 18 possible combinations.
With two bits, there are (2^2) possible combinations, which equals 4. The combinations are: 00, 01, 10, and 11. Each bit can be either 0 or 1, leading to these four distinct configurations.
At least 11 of them.
To calculate the number of possible combinations of the digits 1, 3, 7, and 9, we can use the formula for permutations of a set of objects, which is n! / (n-r)!. In this case, there are 4 digits and we want to find all possible 4-digit combinations, so n=4 and r=4. Therefore, the number of possible combinations is 4! / (4-4)! = 4! / 0! = 4 x 3 x 2 x 1 = 24. So, there are 24 possible combinations using the digits 1, 3, 7, and 9.
The total number of 1-bit combinations is 2. This is because a single bit can have two possible values: 0 or 1. Therefore, the combinations are {0, 1}.