I only know how to do this by complementary counting:
(All 4-digit number) - (All 4-digit numbers with no consecutive digits that are the same)
All 4-digit number:
This is simply 1000 to 9999 inclusive, for a total of 9000 distinct numbers.
All 4-digit numbers with no consecutive digits that are the same:
We can count this through constructive counting. For the thousand digit, we have 9 possibilities (1 to 9, 0 is not an option). For the hundred digit, we have 10 possibilities (0 to 9). However, Since consecutive digits must not be the same, we have now only 9 possibilities (discarding whatever number the thousand digit is). The same goes for the tens and unit digit.
For a total of $9^4 = 6561$ possibilities.
Subtract the two, there are $9000 - 6561 = 2439$ of such numbers.
You made two errors:
- You did not account for the fact that there are only $9$ choices for the leading digit when neither number in the pair of consecutive equal even digits is in the thousands place.
- You have subtracted numbers in which there are three or more consecutive even digits more than once.
First, we observe that there are $9 \cdot 10 \cdot 10 \cdot 10 = 9000$ four-digit positive integers.
Let $A_1$ be the set of four-digit positive integers in which the thousands place and hundreds place contain equal even digits. Let $A_2$ be the set of four-digit positive integers in which the hundreds place and tens place contain equal even digits. Let $A_3$ be the set of four-digit positive integers in which the tens place and units place contain equal even digits. Then $A_1 \cup A_2 \cup A_3$ is the set of four-digit positive integers that do contain consecutive equal even digits. By the Inclusion-Exclusion Principle, the number of four-digit positive integers that do contain consecutive even digits is
$$|A_1 \cup A_2 \cup A_3| = |A_1| + |A_2| + |A_3| - |A_1 \cap A_2| - |A \cap A_3| - |A_2 \cap A_3| + |A_1 \cap A_2 \cap A_3|$$
$|A_1|$: Since we cannot use zero, there are four ways of choosing the even digit that occupies both the thousands and hundreds place. There are ten choices for each of the remaining digits. Hence, $|A_1| = 4 \cdot 10 \cdot 10 = 400$.
$|A_2|$: Since we cannot use zero, the thousands place can be filled in nine ways. There are five ways to choose the even digit that fills both the hundreds and tens places. There are ten ways to fill the units place. Hence, $|A_2| = 9 \cdot 5 \cdot 10 = 450$.
$|A_3|$: We can fill the thousands place in nine ways and the hundreds place in ten ways. There are five ways to choose the even digit that occupies both the tens and units places. Hence, $|A_3| = 9 \cdot 10 \cdot 5 = 450$.
$|A_1 \cap A_2|$: Since we cannot use zero, there are four ways to choose the
even digit that occupies the thousands, hundreds, and tens places. There are ten ways to fill the units place. Hence, $|A_1 \cap A_2| = 4 \cdot 10 = 40$.
$|A_1 \cap A_3|$: There are four ways to choose the even digit that occupies both the thousands and hundreds places and five ways to choose the even digit that occupies both the tens and units places. Hence, $|A_1 \cap A_3| = 4 \cdot 5 = 20$.
$|A_2 \cap A_3|$: There are nine ways to fill the thousands place. There are five ways to choose the even digit that occupies the hundreds, tens, and units places. Hence, $|A_2 \cap A_3| = 9 \cdot 5 = 45$.
$|A_1 \cap A_2 \cap A_3|$: Since we cannot use zero, there are four ways to choose the even digit that occupies the thousands, hundreds, tens, and units places. Hence, $|A_1 \cap A_2 \cap A_3| = 4$.
Therefore, the number of four-digit positive integers that do contain consecutive even equal digits is
\begin{align*}
|A_1 \cup A_2 \cup A_3| & = |A_1| + |A_2| + |A_3| - |A_1 \cap A_2| - |A \cap A_3| - |A_2 \cap A_3| + |A_1 \cap A_2 \cap A_3|\\
& = 400 + 450 + 450 - 40 - 20 - 45 + 4\\
& = 1199
\end{align*}
Therefore, the number of four-digit positive even integers that do not contain consecutive equal even digits is
$9000 - 1199 = 7801$.
Best Answer
You say you have $9000$ four-digit numbers.
To count numbers with no consecutive repeat digits is quite easy: you say you have $9$ choices for the first digit; given the first you have $10-1=9$ choices for the second; given the second you have $9$ choices for the third; given the third you have $9$ choices for the fourth.
$$\dfrac{9^4}{9 \times 10^3} = 0.729$$ so if this is what you were trying to do then you have made a mistake.
(Added) If alternatively you are looking for the numbers which do not have any consecutive $5$s, the easiest way to count is to look at those which do, as you attempted.
There are three possible patterns of the forms
55AA,B55AorCD55whereAis any digit ,Bis any digit except $0$ or $5$,Cis any digit except $0$, andDis any digit except $5$. So there are $10\times 10 + 8 \times 10 + 9\times 9 = 261$ four-digit numbers which have consecutive $5$s and so $9000-261 = 8739$ which do not.$$\dfrac{8739}{9000} = 0.971$$ so again you have an error, but much closer this time. You might try to spot how you have over counted the second two patterns