Following the logic from the link Number of 4-digit numerals with at most 2 different digits
Writing steps in the same manner :
1) In this question, we have to find exactly 4. ( not like in that question where atmost 2 distinct digits were required ) . So this case NA.
2) Here I tried C(9,4)( C(4,2)*6!/2!.2! + C(4,1)*6!/3! ). This is the case where we exclude 0.
3) In this case, we consider 0, I didnt understand this case in that question too.
So , Can anyone verify if my approach is correct. What could be the third case in this question, following like they did in the question linked above ?
OR if you can suggest other approach altogether. I consider few more approaches for the linked question (How many 4 digit numbers are there which contains not more than 2 different digits?). But, I could not extrapolate their concept to 6 digits and 4 distinct numbers.