how many integers are there between 10 000 and 99 999
- a) whose digits are are each odd?
- b) with no repeated digits?
- c) with no repeated digits and whose digits are each odd?
I know there are 90 000 integers. Help please.
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a) Each digit has 5 possibilities, so we have $$5^5 = 3125$$ numbers.
b) The first two digit have 9 possibilities, the third 8, the fourth 7 and the fifth 6. So, we have $$9*9*8*7*6 = 27216$$ numbers.
c) The odd digits 1,3,5,7,9 can be written down in $$5! = 120$$ different ways, so we have 120 numbers.
Hints - since you show no work of your own - this is essentially a counting question, and you need to be systematic:
For the first, how many ways can the first digit be odd, and the second digit, and the third ...
For the second how many sets of five distinct non-zero digits can you choose? How many numbers can you make from each set? How many sets of five distinct digits can you choose if zero is included? How many numbers can you make from each set (zero can't go first).
For the third, how many sets of five distinct odd digits can you choose? [Zero is even] How many numbers can you make from each set?
I believe you should be able to answer these questions and combine the answers to give the numbers you need. Give it a try, and then if you get stuck on some specific point, edit your question to ask about that specific problem.
This method of splitting counting questions into a number of simpler questions is one I have found helpful over the years. There are other ways of approaching them too.