Certainly! Let's solve the problem step by step.
Given:
1. \( r = \binom{3}{1} \)
2. \( s = \binom{-2}{1} \)
3. We need to calculate \( 6(x + 2s) \)
### Step 1: Calculate \( r \)
The binomial coefficient \( \binom{3}{1} \) represents the number of ways to choose 1 element from a set of 3 elements. This can be calculated using the formula for binomial coefficients \( \binom{n}{k} = \frac{n!}{k!(n-k)!} \).
[tex]\[
\binom{3}{1} = \frac{3!}{1!(3-1)!} = \frac{3!}{1! \cdot 2!} = \frac{3 \times 2 \times 1}{1 \times (2 \times 1)} = 3
\][/tex]
So,
[tex]\[
r = 3
\][/tex]
### Step 2: Calculate \( s \)
The binomial coefficient \( \binom{-2}{1} \) usually isn't defined in the classical combinatorial sense because it involves a negative number. However, in the context of combinatorial interpretation with negative values, there is a convention that can be used.
For \( \binom{-n}{k} \), it is often defined as:
[tex]\[
\binom{-n}{k} = (-1)^k \binom{n+k-1}{k}
\][/tex]
In this case, for \( n = 2 \) and \( k = 1 \):
[tex]\[
\binom{-2}{1} = (-1)^1 \binom{2 + 1 - 1}{1} = - \binom{2}{1}
\][/tex]
Now calculating \( \binom{2}{1} \):
[tex]\[
\binom{2}{1} = \frac{2!}{1!(2-1)!} = \frac{2!}{1! \cdot 1!} = \frac{2}{1} = 2
\][/tex]
So,
[tex]\[
\binom{-2}{1} = -2
\][/tex]
Therefore,
[tex]\[
s = -2
\][/tex]
### Step 3: Calculate \( 6(x + 2s) \)
We need to evaluate the expression \( 6(x + 2s) \).
Substitute \( s = -2 \):
[tex]\[
6(x + 2s) = 6(x + 2(-2)) = 6(x - 4)
\][/tex]
### Step 4: Write the final expression
The answer in terms of \( x \) is:
[tex]\[
6(x - 4)
\][/tex]
