4
There are 8*7/(2*1) = 28 combinations.
There are 9C2 = 9*8/(2*1) = 36 2-digit combinations.
There are 8C6 = 8*7/(2*1) = 28 combinations.
8
28, or 56 counting reversals
There are 8*7/(2*1) = 28 combinations.
There are 9C2 = 9*8/(2*1) = 36 2-digit combinations.
8*7/(2*1) = 28
There are 8C6 = 8*7/(2*1) = 28 combinations.
8
28, or 56 counting reversals
To calculate the combinations for the numbers 2, 6, and 8, we need to use the formula for combinations, which is nCr = n! / r!(n-r)!. In this case, we have 3 numbers (n=3) and we want to choose 2 of them (r=2). So, the combinations would be 3C2 = 3! / 2!(3-2)! = 3. Therefore, the combinations for 2, 6, and 8 are (2, 6), (2, 8), and (6, 8).
There are 8!/[6!(8-6)!] = 8*7/2 = 28 - too many to list.
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.
1 byte is 8 bits. That's 8 1s or 0s. 2 bytes is 8*2=16 bits (1s/0s). That is 2^16=65536 possibilities. Therefore, there are 65,536 different combinations with 2 bytes.
There are only four combinations but there are 8 permutations.
Your a bum 4 lising 2 me 2 4 6 8 2 4 8 6 2 6 4 8 2 6 8 4 2 8 4 6 2 8 6 4 4 2 6 8 4 2 8 6 4 6 2 8 4 6 8 2 4 8 2 6 4 8 6 2 6 2 4 8 6 2 8 4 6 4 2 8 6 4 8 2 6 8 2 4 6 8 4 2 8 2 4 6 8 2 6 4 8 4 2 6 8 4 6 2 8 6 2 4 8 6 4 2 That's all 24 possible combinations.