n how many ways can you choose 5 marbles from a bag of 10 marbles if the order does not matter?

To determine the number of ways to choose 5 marbles from a bag of 10 marbles when the order does not matter, we use the concept of combinations. The formula for combinations is given by:

$C(n, k) = \frac{n!}{k!(n-k)!}$

where $n$ is the total number of items, $k$ is the number of items to choose, and $!$ denotes factorial, which is the product of all positive integers up to that number.

In this case, $n = 10$ and $k = 5$. Plugging these values into the formula, we get:

$C(10, 5) = \frac{10!}{5!(10-5)!}$

First, calculate the factorials:

$10! = 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 3,628,800$

$5! = 5 \times 4 \times 3 \times 2 \times 1 = 120$

$(10-5)! = 5! = 120$

Now, substitute these values back into the combination formula:

$C(10, 5) = \frac{3,628,800}{120 \times 120}$

$C(10, 5) = \frac{3,628,800}{14,400}$

$C(10, 5) = 252$

Therefore, there are 252 ways to choose 5 marbles from a bag of 10 marbles when the order does not matter.