To solve the given problem, let's analyze each step in Lorelei's solution and compare it with the correct procedure to calculate [tex]\(\binom{12}{10}\)[/tex], which is the number of ways to choose 10 items from 12.
The given expression is:
[tex]\[
\frac{12!}{(12-10)! \cdot 10!}
\][/tex]
### Step-by-Step Analysis of Lorelei's Solution:
1. Step 1: Subtract within parentheses and simplify
Lorelei wrote: [tex]\(\frac{61}{(2) 151}\)[/tex]. This step seems incorrect and doesn't reflect the standard calculation of factorials and binomial coefficients. The subtraction within the parentheses should be handled in the context of factorials first.
2. Correct Step 1: Calculate the factorials within the binomial coefficient formula.
We need to simplify the expression as follows:
[tex]\[
\frac{12!}{(12-10)! \cdot 10!} = \frac{12!}{2! \cdot 10!}
\][/tex]
3. Step 2: Expand the factorials
Lorelei wrote: [tex]\(\frac{6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}{2 \cdot 1 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}\)[/tex]. This expansion is incorrect. The correct expansion should account for simplifying [tex]\(12!\)[/tex] and [tex]\((12-10)!\)[/tex]:
[tex]\[
\frac{12 \cdot 11 \cdot 10!}{2! \cdot 10!}
\][/tex]
Now, we can cancel the [tex]\(10!\)[/tex] in the numerator and denominator:
[tex]\[
\frac{12 \cdot 11}{2!} = \frac{12 \cdot 11}{2 \cdot 1}
\][/tex]
4. Step 3: Divide out common factors
Lorelei wrote: [tex]\(\frac{6}{2 \cdot 1}\)[/tex]. Correctly dividing out the common factors from previous step:
[tex]\[
\frac{12 \cdot 11}{2 \cdot 1} = \frac{132}{2} = 66
\][/tex]
### Conclusion:
Now, let's evaluate the given statements based on Lorelei's solution:
- Her work is correct. This statement is incorrect since her steps and simplification contained errors.
- Her answer is correct. This statement is incorrect since her answer was 3, but the correct answer is 66.
- In step 1, the subtraction cannot be completed before the factorial of each number is calculated. This statement is correct.
- In step 1, [tex]\(12! \div 10!\)[/tex] is not equivalent to [tex]\(6! \div 5!\)[/tex]. This statement is correct.
- In step 3, the dividing out of common factors was performed incorrectly. This statement is correct.
- There are sixty-six ways to choose ten items from twelve. This statement is correct.
Thus, the statements that describe Lorelei's solution accurately are:
- In step 1, the subtraction cannot be completed before the factorial of each number is calculated.
- In step 1, [tex]\(12!\)[/tex] divided by [tex]\(10!\)[/tex] is not equivalent to [tex]\(6!\)[/tex] divided by [tex]\(5!\)[/tex].
- In step 3, the dividing out of common factors was performed incorrectly.
- There are sixty-six ways to choose ten items from twelve.
