To simplify the expression \(\frac{nC_3 + nP_3}{n^2} - n\), let's break it down step by step:

### Step 1: Express the combinations and permutations
1. **Combinations**: \( nC_3 \) represents the number of ways to choose 3 objects from a set of \( n \) objects. The formula for \( nC_3 \) is:
   \[
   nC_3 = \frac{n!}{3!(n-3)!} = \frac{n(n-1)(n-2)}{6}
   \]

2. **Permutations**: \( nP_3 \) represents the number of ways to arrange 3 objects from a set of \( n \) objects. The formula for \( nP_3 \) is:
   \[
   nP_3 = \frac{n!}{(n-3)!} = n(n-1)(n-2)
   \]

### Step 2: Substitute these expressions into the given expression
\[
\frac{nC_3 + nP_3}{n^2} - n = \frac{\frac{n(n-1)(n-2)}{6} + n(n-1)(n-2)}{n^2} - n
\]

### Step 3: Simplify the expression inside the fraction
The numerator becomes:
\[
\frac{n(n-1)(n-2)}{6} + n(n-1)(n-2) = \frac{n(n-1)(n-2) + 6n(n-1)(n-2)}{6} = \frac{7n(n-1)(n-2)}{6}
\]

So the expression simplifies to:
\[
\frac{\frac{7n(n-1)(n-2)}{6}}{n^2} - n = \frac{7n(n-1)(n-2)}{6n^2} - n
\]

### Step 4: Simplify further
\[
\frac{7(n-1)(n-2)}{6n} - n
\]

Finally, we combine the terms:
\[
\frac{7(n-1)(n-2)}{6n} - \frac{6n^2}{6n} = \frac{7(n-1)(n-2) - 6n^2}{6n}
\]

This is the simplified form of the expression.

Would you like further details or have any questions?

Here are five related questions for practice:
1. Simplify \(\frac{nP_2 + nC_2}{n^2} - n\).
2. What is the value of \(\frac{nP_3}{nC_3}\)?
3. Simplify \(\frac{nC_4 + nP_4}{n^2} - n\).
4. What is the general formula for \(nC_k\)?
5. Simplify \(\frac{nP_k}{n^k}\) for any \(k\).

**Tip:** Remember that combinations count selections where order doesn't matter, while permutations count selections where order does matter.

Learn how to simplify the expression (nc3+np3)/n²-n with a detailed step-by-step solution. This problem covers key mathematical concepts including combinations, permutations, and factorials. Suitable for students in Grades 10-12.

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