A student needs to create a 7-character alpha-numeric passcode (that is, at least one letter and one digit must be included). She uses lower case letters from her name "venessa see" and digits 1 to 9.

1. Identify the number of available distinct letters and digits.

Distinct letters in "venessa see": $\{v, e, n, s, a\}$ (5 letters)
Distinct digits from 1 to 9: $\{1, 2, 3, 4, 5, 6, 7, 8, 9\}$ (9 digits)
Total distinct characters available = $5 + 9 = 14$

2. Determine the number of choices for the restricted first and last positions.

First position (must be a letter): $5$ choices
Last position (must be a digit): $9$ choices
Number of characters remaining after filling the first and last positions = $14 - 2 = 12$

3. Determine the number of ways to arrange characters in the remaining positions and calculate the total number of passcodes.

Remaining positions to fill = $7 - 2 = 5$
Ways to arrange the remaining 12 characters into 5 positions = $^{12}P_5 = 95,040$
Total number of passcodes = $5 \times 9 \times {}^{12}P_5$
$= 45 \times 95,040 = 4,276,800$

1. Determine the possible patterns to ensure all letters and digits are separated.

For all letters (L) and digits (D) to be separated, they must strictly alternate. Since the passcode is 7 characters long, there are two possible patterns:

1. L-D-L-D-L-D-L (4 letters, 3 digits)
2. D-L-D-L-D-L-D (3 letters, 4 digits)

2. Calculate the number of ways for the `L-D-L-D-L-D-L` pattern.

There are $5$ distinct letters (v, e, n, s, a) and $9$ distinct digits.
Choosing and arranging 4 letters and 3 digits:
$\text{Ways} = {}^5P_4 \times {}^9P_3$
$= 120 \times 504 = 60480$

3. Calculate the number of ways for the `D-L-D-L-D-L-D` pattern and find the total.

Choosing and arranging 3 letters and 4 digits:
$\text{Ways} = {}^5P_3 \times {}^9P_4$
$= 60 \times 3024 = 181440$

Total number of passcodes $= 60480 + 181440 = 241920$

1. Identify the number of choices for the 3 identical letters and determine the available pool for the remaining characters.

The letters in "venessa see" have the following frequencies: v(1), e(4), n(1), s(3), a(1).
To have exactly 3 identical letters, the chosen letter must appear at least 3 times in the name.
Thus, there are $\binom{2}{1} = 2$ choices for this letter ('e' or 's').

Once the identical letters are chosen, we need 4 more distinct characters to form a 7-character passcode.
The pool of available distinct characters consists of the remaining $4$ distinct letters and the $9$ digits (1 to 9).
Total characters in the remaining pool $= 4 + 9 = 13$.

2. Calculate the number of valid ways to select the remaining 4 distinct characters, ensuring at least one digit is included.

We need to select 4 distinct characters from the 13 available.
Total unrestricted ways to select 4 characters $= \binom{13}{4} = 715$.

The passcode must be alpha-numeric. Since we already have 3 letters, we must ensure at least 1 digit is included. The only invalid selection is choosing 4 letters (0 digits).
Number of ways to select 4 letters from the 4 available letters $= \binom{4}{4} = 1$.

Valid selections for the 4 distinct characters $= 715 - 1 = 714$.

3. Determine the number of arrangements for the 7 characters and calculate the total number of passcodes.

For any chosen set of 7 characters (which consists of 3 identical letters and 4 distinct characters), the number of distinguishable arrangements is $\frac{7!}{3!}$.

Total number of passcodes $= 2 \times 714 \times \frac{7!}{3!}$
$= 1428 \times 840$
$= 1,199,520$