How many 5-digit numbers can be formed using digits 2,0,9 at least once?
Is this correct:
No. of ways in which they don't appear: $7*7*7*7*7$ Total ways without restriction: $9*10*10*10*10$ No. of ways in which they appear at least once or as many times: $9*10^4-7^5$
If not then please correct and give your solution.
Let $A_0, A_2, A_9$ represent the sets of five digit numbers which contain at least one $0$, at least one $2$, and at least one $9$ respectively.
Your answer is correct for counting $|A_0\cup A_2\cup A_9|$, the number of five digit numbers which contain at least one of the numbers $0,2,9$ but it need not contain all of the numbers. This would allow for numbers like $22222$ to be considered as "valid" numbers which we wish to count despite not containing any $0$ or $9$.
My interpretation of the question however is that we wish to count $|A_0\cap A_2\cap A_9|$, that is to say, we are wanting to count numbers which contain at least one of each of the numbers $0,2,9$.
Letting $\Omega$ be the set of five digit numbers, approach via inclusion-exclusion. $|A_0\cap A_2\cap A_9|=|\Omega\setminus (A_0^c\cup A_2^c\cup A_9^c)|$
$=|\Omega|-|A_0^c|-|A_2^c|-|A_9^c|+|A_0^c\cap A_2^c|+|A_0^c\cap A_9^c|+|A_2^c\cap A_9^c|-|A_0^c\cap A_2^c\cap A_9^c|$
Calculate each term individually like you would expect, remembering to take extra care about the usage of zeroes that they cannot occur as the first digit in a five digit number.
For example, $|A_2^c\cap A_9^c|$ is the set of five digit numbers with no $2$ and no $9$, of which by multiplication principle there are $7\cdot 8^4$ such numbers while $|A_0^c\cap A_9^c|=8^5$ instead