Observe the following series carefully and answer the question that follows:
$$\text{C \$ 1 \% E 7 2 1 ! @ I * \# @ A 7 * \# > 8 U >}$$

How many numbers are followed by an alphabet if the second half of the series is reversed?

Step 1: Count the total number of elements in the given series.
Elements: \text{C(1), \ (2), 1(3), \%(4), E(5), 7(6), 2(7), 1(8), !(9), @(10), I(11), (12), \#(13), @(14), A(15), 7(16), (17), \#(18), >(19), 8(20), U(21), >(22)}$Totalelements=$ 22$.Step2:Splittheseriesintotwohalves.\ast Firsthalf($ 11$elements):$ \text{C \1 \% E 7 2 1 ! @ I}
* Second half ($11$ elements): $\text{ * \# @ A 7 * \# > 8 U >}$

Step 3: Reverse the second half.

* Reversed second half: $\text{> U 8 > \# * 7 A @ \# *}$

Step 4: Reconstruct the full sequence.

$$\text{C \$ 1 \% E 7 2 1 ! @ I > U 8 > \# * 7 A @ \# *}$$

Step 5: Count how many numbers are immediately followed by an alphabet (Pattern: Number $\to$ Alphabet).
Let's inspect the numbers in the new sequence:

* $1$ (followed by $\%$) $\to$ No
* $7$ (followed by $2$) $\to$ No
* $2$ (followed by $1$) $\to$ No
* $1$ (followed by $!$) $\to$ No
* $8$ (followed by $>$) $\to$ No
* $7$ (followed by $\text{A}$) $\to$ Yes ($7$ is a number, $\text{A}$ is an alphabet).

There is exactly $1$ instance ($\text{7A}$) where a number is followed by an alphabet.