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To deeply understand permutations, first grasp the concept of *factorials*. The factorial of a number, denoted as \(n!\), is the product of all positive integers up to that number. For example, \(6!\) (read as 'six factorial') is calculated as follows:
\[6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 720\]
Factorials are fundamental in counting and permutations, where the factorial of a number gives the total ways to arrange that many distinct items. For instance, if you have 3 items A, B, and C, the number of different ways to arrange them is \(3!\):
\[3! = 3 \times 2 \times 1 = 6\]
Total possible arrangements would be ABC, ACB, BAC, BCA, CAB, CBA. Factorials grow extremely fast with larger numbers, highlighting the immense number of arrangements possible with even a small set.

Permutations are arrangements of objects where order matters. The formula to calculate permutations for 'n' objects taken 'r' at a time is:
\[nPr = \frac{n!}{(n-r)!}\]
However, if you're arranging all 'n' objects (i.e., 'n' is taken 'n' at a time), the formula simplifies to just \(n!\).
For example, to find permutations for 6 letters, we use:
\[6P6 = 6! = 720\]
This is because all 6 letters can be arranged in 720 distinct ways without any repetitions.
Understanding this formula helps a lot in solving permutation problems effortlessly.

Now let’s apply these concepts to a problem involving the arrangement of letters. Let's consider the exercise: 'How many permutations of {a, b, c, d, e, f, g} end with 'a'?'.
First, we fix 'a' at the last position. This leaves us with the task of arranging the remaining 6 letters: {b, c, d, e, f, g}.
The number of permutations of these remaining letters is simply \(6!\):
\[6! = 720\]
Thus, there are 720 ways to arrange the set of letters such that 'a' is at the end. If a problem constrains an element to a fixed position, reduce your focus to the remaining items. Arrangements might seem tricky without segmenting the problem, but breaking it down always simplifies it.