Answer
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Step 1:: Understand the problem
The problem asks us to find the value of the combination $${9 \choose 2}$$.
A combination is a selection of items from a larger set, where the order of the items does not matter.
Step 2:: Recall the combination formula
where $$n$$ is the total number of items, $$k$$ is the number of items to choose, and "!" denotes factorial.
The formula for combinations is given by:
Step 3:: Plug in the values into the formula
{9 \choose 2} = \frac{9!}{2!(9-2)!}
Plugging these values into the formula, we get:
Step 4:: Simplify the expression
{9 \choose 2} = \frac{9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{2 \times 1 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}
First, let's simplify the factorials: Now, plug these simplified factorials into the original expression:
Step 5:: Cancel out common factors
{9 \choose 2} = \frac{9 \times 8}{2 \times 1}
Cancel out the common factors in the numerator and the denominator:
Step 6:: Perform the division
{9 \choose 2} = 9 \times 4
Divide the numbers in the numerator by the numbers in the denominator:
Step 7:: Multiply the numbers
Multiply the numbers to get the
Final Answer
{9 \choose 2} = 36
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