To find the number of ways to choose 5 students from a class of 17, we use combinations. The combination formula is given by:
\[ C(n, k) = \frac{n!}{k!(n-k)!} \]
where \(n\) is the total number of students, \(k\) is the number of students to choose, and \(n!\) (n factorial) is the product of all positive integers up to \(n\).
In this case, \(n = 17\) and \(k = 5\). Plugging these values into the formula:
\[ C(17, 5) = \frac{17!}{5!(17-5)!} = \frac{17!}{5! \cdot 12!} \]
We can simplify this by calculating just the necessary top part of the factorial:
\[ C(17, 5) = \frac{17 \times 16 \times 15 \times 14 \times 13}{5 \times 4 \times 3 \times 2 \times 1} \]
Calculating the numerator:
\[ 17 \times 16 = 272 \] \[ 272 \times 15 = 4080 \] \[ 4080 \times 14 = 57120 \] \[ 57120 \times 13 = 742560 \]
Now the numerator is 742560. Now calculating the denominator:
\[ 5 \times 4 \times 3 \times 2 \times 1 = 120 \]
Now divide the two results:
\[ C(17, 5) = \frac{742560}{120} = 6188 \]
Thus, the number of ways to choose 5 students from a class of 17 is:
6,188
So the correct response is:
6,188