You are selecting an 8-character password using 26 letters and numbers 0 through 9. In how many ways could your password contain a) at least two letters b) at least two numbers c) at least two letters and two numbers

This is a question involve exponents and summations. The total number of possibilities for all 8 digits with no restrictions is as follows:
Assuming that repetition is allowed, we have (26+10)^8 = 2.8 * 10^12 possibilities. Now for the limitations, consider the possibilities as follows:
10^0∗26^8=208,827,064,576
10^1∗26^7=642,544,814,080
10^2∗26^6=864,964,172,800
10^3∗26^5=665,357,056,000
10^4∗26^4=319,883,200,000
10^5∗26^3=98,425,600,000
10^6∗26^2=18,928,000,00
10^7∗26^1=2,080,000,000
10^8∗26^0=100,000,000
Now for the limitations: If the password must include two letters, then we subtract the possibilities that involve fewer than two letters (100,000,000+2,080,000,000) so the answer is 2.8 x 10^12.
For part b.) we subtract the options that don't include two numbers (208,827,064,576+642,544,814,080) so the final answer is 1,969,738,028,800.
For part c.) we subtract both subsets involving possibilities without at least two of both numbers and letters (100,000,000+2,080,000,00) and (208,827,064,576+642,544,814,080) so the answer is 1.9 X 10^2.