Let's solve each part step by step.

### Part a: \( 44 \times 45 + 46 + 44 + 9 \times 44 \)

First, we simplify the multiplication terms:
\[ 44 \times 45 = 1980 \]
\[ 9 \times 44 = 396 \]

Now, we add all the terms together:
\[ 1980 + 46 + 44 + 396 \]

Adding them step by step:
\[ 1980 + 46 = 2026 \]
\[ 2026 + 44 = 2070 \]
\[ 2070 + 396 = 2466 \]

So, the result for part a is:
\[ 2466 \]

### Part b: \( 2 \times 31 \times 12 + 4 \times 6 \times 42 + 8 \times 27 \times 3 \)

First, we simplify the multiplication terms:
\[ 2 \times 31 \times 12 = 744 \]
\[ 4 \times 6 \times 42 = 1008 \]
\[ 8 \times 27 \times 3 = 648 \]

Now, we add all the terms together:
\[ 744 + 1008 + 648 \]

Adding them step by step:
\[ 744 + 1008 = 1752 \]
\[ 1752 + 648 = 2400 \]

So, the result for part b is:
\[ 2400 \]

### Part c: \( 11 + 13 + 15 + \ldots + 25 \)

This is an arithmetic sequence where the first term \( a = 11 \) and the common difference \( d = 2 \). We need to find the sum of this sequence up to the term 25.

First, find the number of terms \( n \) in the sequence. The \( n \)-th term of an arithmetic sequence is given by:
\[ a_n = a + (n-1)d \]

Set \( a_n = 25 \):
\[ 25 = 11 + (n-1) \times 2 \]
\[ 25 = 11 + 2n - 2 \]
\[ 25 = 9 + 2n \]
\[ 16 = 2n \]
\[ n = 8 \]

Now, use the sum formula for an arithmetic sequence:
\[ S_n = \frac{n}{2} \times (a + l) \]
where \( l \) is the last term.

Here, \( n = 8 \), \( a = 11 \), and \( l = 25 \):
\[ S_8 = \frac{8}{2} \times (11 + 25) \]
\[ S_8 = 4 \times 36 \]
\[ S_8 = 144 \]

So, the result for part c is:
\[ 144 \]

### Summary:
a. \( 2466 \)
b. \( 2400 \)
c. \( 144 \)