00:01
Starting with part a, where all characters must be in lowercase, that is 26 letters because of a to z in the alphabet.
00:12
So the number of permutations, or p, would be 26 to the power of 10, which equals 1 .007 times 10 to the power of 15.
00:25
For part b, each character must alternate between uppercase and digit, and each occurrence has an uppercase letter, has 25 remaining options, excluding the one used previously, and each occurrence has a digit of 9 remaining options, also excluding the one used.
00:55
So the permutation would be 36 times 25 times 9 to the power of 4, which equals 2 .38 times 10 to the power of 12.
01:10
For part c, the permutation is to arrange 10 distinct lowercase letters alphabetically, and this permutation is equivalent to 10, or 3 ,628 ,800.
01:33
800...