How to calculate number of favourable cases for the following problem How to find the total cases where a four digit number has two like digits,three like digits, two pairs of like digits, four like digits etc...
 A: I will assume that the four-digit numbers start with $1000$. If you allow things like $0075$ as four-digit numbers, then the calculations become substantially smoother. 
I am sure that you can answer some of the questions easily. For example, one can list and count the "four like digit" numbers. They are $1111$, $2222$, and so on up to $9999$, a total of $9$.
Probably the next easiest to count is the "two pairs" case. The first digit can be chosen in $9$ ways. Suppose that for example we choose the first digit to be $4$. There are $3$ ways to decide where the other $4$ will go. And once we have decided that, there are $9$ choices for what digit the remaining two "holes" will be filled by. So we end up with $(9)(3)(9)$ "two pairs" numbers.
We now count the "one pair" numbers. It is maybe easiest to break up the work into two cases: (i) The pair involves the first digit and (ii) The pair does not involve the first digit. 
For (i), the first digit can be chosen in $9$ ways. the location where the first digit is matched can then be chosen in $3$ ways. Once we have done that, we have two empty slots left. We can fill the leftmost one in $9$ ways. For each of these, we can fill the remaining empty slot in $8$ ways, for a total of $(9)(3)(9)(8)$.  For (ii), the first digit can be chosen in $9$ ways. For each such choice, the slot that will not hold the pair can be chosen in $3$ ways, and the number that fills that slot can then be chosen in $9$ ways. Once we have done that, the number we have two of can be chosen in $8$ ways, for a total of $(9)(3)(9)(8)$. Thus the number of "one pair" numbers is $(9)(3)(9)(8)+(9)(3)(9)(8)$. 
Finally, we deal with the "one triple" numbers. The triple could (i) Not involve the first digit or (ii) Involve the first digit.  For (i), the first digit can be chosen in $9$ ways, and then the digit we have $3$ of can be chosen in $9$ ways, for a total of $(9)(9)$. For (ii), the first digit can be chosen in $9$ ways. For each such choice, the location of the odd digit can be chosen in $3$ ways, and then the odd digit can be chosen in $9$ ways, for a total of $(9)(3)(9)$. thus the number of "one triple" numbers is $(9)(9)+(9)(3)(9)$. 
Remark: There are many other ways to organize the calculation. One clever thing to do would be to temporarily allow $0$ as an initial digit: then the calculations are smoother. After doing the computations, we subtract the forbidden cases. I would urge you to experiment with ways of counting. The numbers I have obtained can then serve as a check.   
A: For this kind of counting argument, I like to use 
multinomial coefficients to keep things straight. 
Suppose I want to count the number of ordered 
strings of length four, taken from a set of ten digits,
 showing two different pairs; like this $0101$, $6446$, etc. 
Note that I am not going with the everyday meaning of four-digit number.
I first count the number of ways to select the digits,
 then multiply by the number of ways to arrange the digits.  
$$
\begin{array}{cc}
\text{Choose digits} & \text{Arrange them}\\[5pt]
{10\choose 8,0,2}&{4\choose 2,2}
\end{array}
$$
This works out to $270$. 
How do you read those multinomial 
coefficients? The first one gives the frequencies of the ten digits:
 there will be 8 digits that do not appear, 0 digits that appear exactly once, 
and 2 digits that appear twice. For the second one, under the four we have 
 the pattern 2,2; that is, two pairs. 
You can change the numbers to solve similar problems.
Let's try 3 pairs in a string of length 6:
$${10\choose 7,0,3}\times {6\choose 2,2,2}=10800. $$
How about one triple, one pair, and one single in a string of length 6:
$${10\choose 7,1,1,1}\times{6\choose 1,2,3}=43200. $$  
One last example: If you roll a die ten times, how many outcomes have one quintuple, two pairs, and a single?
$${6\choose 2,1,2,0,0,1}\times {10\choose 1,2,2,5}=1360800. $$  
