It is easier to count the total directly: you have five possibilities for the first digit (any digit except $0$), and six each for the remaining three digits, so the total number is $5\times 6^3$. This agrees with your count ($6^4-6^3 = 6^3(6-1) = 6^3\times 5$), but I think it's easier to just count them directly.
You can use the same method for counting the possibilities of the first two digits: five possibilities for the first digit, six for the second, for a total of $5\times 6$; this is the same as your count, $6^2-6^1 = 6(6-1)$.
The rest is almost correct; again, you can be a bit briefer with the count of the last two digits: the first digit can be any of the six possibilities. If the first digit is 0, 2, or 4, then the second digit must be 0 or 4 to get a multiple of $4$; so for each of the three possibilities you have two $2$-digit combinations, giving $3\times 2=6$ possibilities. If the first digit is $1$, $3$, or $5$, then the second digit must be $2$, so you have only three more. The total here is the sum of the two, so we have $6+3=9$ possibilities.
What you forgot is that $00$ also works.
Since the choices for the first two digits are independent of the choices for the last two, so you multiply the totals. The total will then be
$$5\times 6 \times 9 = 270.$$
The fact that the total number (1080) is larger than the count of those that are divisible by 4 should not be a surprise: there are lots of numbers among the 1080 4-digit numbers in which every digit is one of 0, 1, 2, 3, 4, and 5 that are not divisible by $4$. So I'm not sure what your last line is meant to represent.
In my experience, if it is difficult to calculate a probability, it is sometimes easier to calculate the inverse.
So how many 10 digit strings do not contain consecutive 0s?
We can ignore the first digit since one digit in of itself will have no impact on the probability.
From there, the probability that a single digit does not cause that string to contain consecutive 0s is inverse of the the probability that both the previous and the current digits are 0.
So if we were solving for 2 digit strings, the probability would simply be the inverse of the probability of having 0 digit followed by another 0 digit, or $1 - (\frac{1}{10\times10})$.
If we were to then apply this to a 3 digit string, the probability would be compounded with the inverse of the probability of the last digits having 0. This simply means to find the probability of not finding 0s in the first two digits and not finding 0s in the second two digits. This also covers the case of both finding digits in the first 2 digits as well as finding digits in the last 2 digits.
The probability of the first two digits not both having 0 is $1 - (\frac{1}{10\times10})$, and the probability of the last two digits not both having 0 is $1 - (\frac{1}{10\times10})$. Compounded, you would get $$1 - \frac{1}{\frac{1}{1 - (\frac{1}{10\times10})}\times\frac{1}{1 - (\frac{1}{10\times10})}}$$ Simplying a bit, we get:
$$1 - (1 -\frac{1}{10\times10})^{2}$$
Why the 2? Because we have 3 digits and 2 pairs of digits to consider. Generalizing, we would have:
$$1 - (1 - \frac{1}{10\times10})^{n-1}$$
Where $n$ is the number of digits. Take this probability and multiply times $10^n$ or in this case $10^{10}$ (total number of possible combinations) and you get the number of digits containing consecutive zeros.
It would seem that you're making the mistake of assuming the pattern is linear. When you're finding all configurations of non-0 digits in the case in which m is 1, there should be just as many configurations of 0 digits in the case in which m is 9. So:
$$xxxxxxxxx0,0xxxxxxxxx$$
If your linear calculation were accurate, then there should be 10000000000 ways to place that singular 0 digit! Hope that helps!
Best Answer
If I am not getting your question wrong then your answers are wrong.
a)If a 4 digit number contains exactly four 9's , then $^9C_4$ is not your solution , there is only one way i.e 9999 , so answer should be 1
b)This one is not as simple as you did,exactly two same will have conditions either $1^{st}$ and $2^{nd}$ are same or $1^{st}$ and $3^{rd}$ are same and so on , there will be 6 such conditions , you will have to solve them separately , .
c)It's nowhere mentioned that digits other than last one cannot be even so answer should be 10 x 10 x 10 x 3 , for (4,6,8 viz. even digits greater than 2)