30
The sum you require is 4!/2!2!, which is 6. This means there are 6 different two-digit numbers. * * * * * Wrong! The above formula is for the number of combinations, not permutations. However, since the number 23 is different from the number 32, you require permutations and not combinations. There are 4*3 = 12 23, 25, 27, 32, 35, 37, 52, 53, 57, 72, 73, 75.
The answer is 15C6 = 15*14*13*12*11*10/(6*5*4*3*2*1) = 5,005
There is only one possible combination of a 13 digit number created from 13 digits. In a combination, the order of the digits does not matter so that 123 is the same as 132 or 312 etc. If there are 13 different digits (characters) there is 1 combination of 13 digits 13 combinations of 1 or of 12 digits 78 combinations of 2 or of 11 digits and so on There are 213 - 1 = 8191 in all. If the characters are not all different it is necessary to have more information.
Te probability of rolling a sum of 7 with two fair dice is 6 in 36, or 1 in 6, or about 0.1667.Of all the possible combinations of two dice, the sum of 7 has the highest probability, with the other combinations decreasing down to 2 and 12, with probabilities of 1 in 36, or about 0.0278.
12
There are 24C12 = 24!/[12!*12!] = 2,704,156 combinations.
12 Combinations.
12 bits
8.796 x 10^12 bits
There are 24C12 = 24*23*...*13/(12*11*...*1) = 2,704,156 combinations.
You would get 4!/2! = 12 combinations.
12 Bits
2^12=4096
12
Only three: 12, 13 and 23. Remember that the combinations 12 and 21 are the same.
12 bits. Thus the reason the file allocation table is called FAT12.