How many five digit integers (base-10 numeral system); a) start with a $9$ ? b) contain a $9$ ? c) do not contain a $9$? How many five digit integers (base-10 numeral system)
a) start with a $9$? 
b) contain a $9$? 
c) do not contain a $9$?
For a) I think its $10000$; we already have the first number $9$ and we need to find the rest of $4$ numbers. Each number has $10$ ways so $10 \times 10 \times 10 \times 10 = 10000$.
Part b and c I don't understand. Please explain.
 A: A) start with 9:  $1\times 10\times 10\times 10\times 10 = 10^4 = 10000$.  (One choice for the first digit ten for all the others)
C) has no 9s:  $8\times 9\times 9\times 9\times 9 = 8\times 9^4 = 52488$.  (Nine choices for each of the digits except the first that can't be zero).
Z) how many five digits numbers total are there:  $9\times 10\times 10\times 10\times 10 = 9\times 10^5 = 90000$.  (9 for the first digit as we cant start with 0.  Well... we can and we'd have 100000 and that'd be fine but then I'd have to deal with nitpickers and you will have questions where leading zeros will be excluded so we'll deal with those now)
B1) How many with at least 1 nine:  Z - C = $90000 - 52488 = 37512$.
B2) How many with exactly 1 nine: Start with 9 + Second is 9 + .... + ends with 9 = $1\times 10\times 10\times 10\times 10 + 9\times 1\times 10\times 10\times 10 + ... 9\times 10\times 10\times 10\times 1 = 10000 + 4\times 9\times 10\times 10\times 10 = 36000$
A: (a) is easy. The first digit must be 9, then 10 choices for each of the other digits, so $10000$ possibilities.
(b) Look at it this way: the first 9 could be in place 1, place 2, place 3 ... We have already found the answer for place 1: 10000. If it is in place 2, then we have 8 choices for the 1st digit (not 9 and not 0), 1 for the 2nd (must be 9), 10 for each of the others, total 8000. If it is in place 3, then we have 8 for the first, 9 for the second, one for the third, and 10 for each of the last two, total 7200. Similarly if it is in place 4 we have 6480 possibilities. If it is in the last place, then we have 5832 possibilities. Total 37512.
(c) The first digit can be anything but 0 or 9, each of the others can be anything except 9, so $8\cdot 9^4=52488$ possibilities.
A: Hint:  It's easier if you do part (c) before doing part (b), methinks.  Because once you find the answer to (c), everything else is (b).
(Note I'm taking (b) to mean "contain one or more 9s" because of its parallel wording with (c) -- as in, (c) appears to be the complement set of (b).)
