So, we need to find the value of $7C0$:
$$7C0 = \frac{7!}{0!(7-0)!}$$
Now, we know that $0! = 1$ and $7! = 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1$. So, we can plug these values into the formula:
$$7C0 = \frac{7!}{0!(7!)} = \frac{7!}{1 \times
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