Determine the number of combinations (subsets) of the following.
9 objects taken 5 at a time
To determine the number of combinations of 9 objects taken 5 at a time, we use the formula for combinations:
\[ C(n, k) = \frac{n!}{k!(n-k)!} \]
Where:
- \( n \) is the total number of objects (9 in this case),
- \( k \) is the number of objects to choose (5 in this case),
- \( ! \) denotes factorial, which is the product of all positive integers up to that number. For example, \( 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \).
So, for our specific case,
\[ C(9, 5) = \frac{9!}{5!(9-5)!} \]
Now we calculate the factorial values:
\[ 9! = 9 \times 8 \times 7 \times 6 \times 5! \]
\[ \text{(since we'll cancel out } 5! \text{, we don't need to calculate it further)} \]
\[ 4! = 4 \times 3 \times 2 \times 1 = 24 \]
Now we insert these values into the combination formula:
\[ C(9, 5) = \frac{9 \times 8 \times 7 \times 6 \times 5!}{5! \times 24} \]
The \( 5! \) in the numerator and denominator cancel each other out:
\[ C(9, 5) = \frac{9 \times 8 \times 7 \times 6}{24} \]
Now, divide by 24 (which is \( 4 \times 6 \)):
\[ C(9, 5) = \frac{9 \times 8 \times 7}{4} \]
Dividing \( 8 \) by \( 4 \), we get:
\[ C(9, 5) = 9 \times 2 \times 7 \]
\[ C(9, 5) = 18 \times 7 \]
\[ C(9, 5) = 126 \]
So there are 126 different ways to choose 5 objects from a set of 9.