To evaluate the expression:

$\frac{11P2}{2!} - 11C2$

Let's break it down into two parts:

1. Permutations ( $11P2$ ):

The permutation formula for $nP r$ is:

$nP r = \frac{n!}{(n-r)!}$

For $11P2$, we have:

$11P2 = \frac{11!}{(11-2)!} = \frac{11!}{9!}$

Now simplify $11!$ and $9!$:

$11! = 11 \times 10 \times 9!$

So,

$11P2 = \frac{11 \times 10 \times 9!}{9!} = 11 \times 10 = 110$

2. Factorial of 2 ( $2!$ ):

$2! = 2 \times 1 = 2$

Now, compute the first part of the expression:

$\frac{11P2}{2!} = \frac{110}{2} = 55$

3. Combinations ( $11C2$ ):

The combination formula for $nC r$ is:

$nC r = \frac{n!}{r!(n-r)!}$

For $11C2$, we have:

$11C2 = \frac{11!}{2!(11-2)!} = \frac{11!}{2!9!}$

Using $11! = 11 \times 10 \times 9!$ and $9!$ cancels out:

$11C2 = \frac{11 \times 10}{2!} = \frac{110}{2} = 55$

Final Calculation:

Now substitute these values back into the original expression:

$\frac{11P2}{2!} - 11C2 = 55 - 55 = 0$

Thus, the value of the expression is:

$\boxed{0}$

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Would you like further details or have any questions? Here's a list of related topics you might find interesting:

1. How does the permutation formula change with different values of $r$?
2. The difference between permutations and combinations in real-life examples.
3. How do factorials behave as the numbers increase?
4. Application of combinations and permutations in probability theory.
5. How do you derive the combination formula?

Tip: Always be cautious when interpreting permutations and combinations — permutations consider order, while combinations do not.