To evaluate the expression $\frac{{11C7}}{{11P3}}$, we need to first compute $11C7$ and $11P3$.

Step 1: Compute $11C7$

The combination formula $nCk$ is given by:

$nCk = \frac{{n!}}{{k!(n-k)!}}$

For $11C7$, we substitute $n = 11$ and $k = 7$:

$11C7 = \frac{{11!}}{{7!(11-7)!}} = \frac{{11!}}{{7!4!}}$

Now, calculating the factorials:

$11! = 11 \times 10 \times 9 \times 8 \times 7!$

The $7!$ cancels out:

$11C7 = \frac{{11 \times 10 \times 9 \times 8}}{{4!}} = \frac{{11 \times 10 \times 9 \times 8}}{{24}} = \frac{{7920}}{{24}} = 330$

So, $11C7 = 330$.

Step 2: Compute $11P3$

The permutation formula $nPk$ is given by:

$nPk = \frac{{n!}}{{(n-k)!}}$

For $11P3$, we substitute $n = 11$ and $k = 3$:

$11P3 = \frac{{11!}}{{(11-3)!}} = \frac{{11!}}{{8!}}$

Now, calculating the factorials:

$11! = 11 \times 10 \times 9 \times 8!$

The $8!$ cancels out:

$11P3 = 11 \times 10 \times 9 = 990$

So, $11P3 = 990$.

Step 3: Compute the final expression

Now we substitute the values of $11C7$ and $11P3$:

$\frac{{11C7}}{{11P3}} = \frac{{330}}{{990}} = \frac{1}{3} \approx 0.3333$

Thus, the value of the expression is $\frac{1}{3}$ or $0.3333$.

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Do you want more details or have any questions?

Here are 5 related questions you might find interesting:

1. What is the difference between combinations and permutations?
2. How would the result change if the expression was $\frac{12C8}{12P4}$?
3. Can you explain how factorials play a role in counting combinations and permutations?
4. What is $11C4$ and how does it relate to $11C7$?
5. How can you generalize the combination and permutation formulas for any $n$ and $k$?

Tip: When working with combinations and permutations, simplify factorials by canceling common terms to make calculations easier!