# Find $$$C{\left(53,6 \right)}$$$

### Your Input

**Find the number of combinations without repetitions $$$C{\left(53,6 \right)}$$$.**

### Solution

The formula is $$$C{\left(n,r \right)} = \frac{n!}{r! \left(n - r\right)!}$$$.

We have that $$$n = 53$$$ and $$$r = 6$$$.

Thus, $$$C{\left(53,6 \right)} = \frac{53!}{6! \left(53 - 6\right)!} = 22957480$$$ (for calculating the factorial, see factorial calculator).

### Answer

**$$$C{\left(53,6 \right)} = 22957480$$$**