Exactly four different on 6 means that there are some repetitions. Maybe one digit repeated 3 times or maybe 2 repeated two times, example: 111235, 224456. So you want to know these two type of variations and sum them.
There are named variations with repetition and you calculate them in two steps:
1) Think the different digits as groups so a number as AABCDA have 4 different groups: A, B, C and D. Over these 4 groups calculate the number of permutations with repetition with the multinomial coefficient
$$PR_{n}^{k_1,k_2,...,k_n}=\binom{n}{k_1,k_2,...,k_n}=\frac{n!}{k_1!k_2!\cdots k_n!};\ n=k_1+k_2+...+k_n$$
In you case you have two different setups as said above so
$$PR_1=\binom{6}{2,2,1,1}=\frac{2\cdot3\cdot4\cdot5\cdot6}{2\cdot2}=180\\
PR_2=\binom{6}{3,1,1,1}=\frac{2\cdot3\cdot4\cdot5\cdot6}{2\cdot3}=120$$
2) Now for each type of permutation you must calculate the number of variations because A, B, C or D could be any digit between 0 and 9. There are variations of 10 elements over 4 positions, so
$$V_n^k=(n)_k=n(n-1)(n-2)\cdots(n-k+1)\\
V=(10)_4=10\cdot9\cdot8\cdot7=5040$$
3) The total of different number for each setup will be $\sum_{i}PR_i\cdot V$ but we need to discount the combinations that start with zero(s) that are a 10% of the total because exist the same probability for any number in any position and they are 10 different numbers, so the probability that the first digit will be a zero is $\frac{1}{10}$.
And the amount $PR_1$ must be divided by two because the two groups of two elements will repeat the same configuration twice, i.e., we have (by example) the configuration of permutations AABBCD where A, B, C and D take values from 0 to 9 (that is counted on the variations) but there is a symmetric permutation as BBAACD that can repeat the same numbers. The same happen for C and D so we must divide again by two to fix the duplicities. These numbers are the permutations of each type of indistinguishable multiplicity, i.e, $2!=2$.
In a similar way happen with the second setup $PR_2$ with B, C and D. So we need to divide between the number of permutations of indistinguishable multiplicities that is $3!=6$ (and $1!=1$).
So the total of cases will be
$$C=\left(PR_1\frac{1}{4}+PR_2\frac{1}{6}\right)V\cdot\frac{9}{10}=(45+20)81\cdot 56=294840$$
The first part is simple. except zero, all digits can go at first and the remaining positions can be either of the six digits. So the total number of 4-digit numbers is:
$$A=5\times6\times6\times6$$
For the second part, the number of 4-digit numbers with no repeat is
$$B=5\times5\times4\times3$$
So the number of them with at least one repeat is $A-B$.
Best Answer
I don't see how the last line evaluates to $64800$, it's $57600$. However, $64800$ is the correct answer. I've giving the calculations below.
First calculate all possible combinations of $3$ odd digits and $3$ even digits, without using a digit more than once. This includes the cases which start with $0$. $$\binom{6}{3}\times\bigg\{\binom{5}{3}\times3!\bigg\}\times\bigg\{\binom{5}{3}\times3!\bigg\}=72000$$ Now calculate the number of such numbers beginning with $0$. $$\binom{5}{3}\times\bigg\{\binom{5}{3}\times3!\bigg\}\times\bigg\{\binom{4}{2}\times2!\bigg\}=7200$$ Subtracting you get the total number of six-digited numbers with $3$ even and $3$ odd digits as $64800$.