Arithmetic of a combinations formula

Question

I am trying to study, and I'm not quite sure how:

$$ \binom{5}{3} \cdot \binom{7}{3} = 350 $$

From my understanding the formula is

$$ \binom{n}{r} = \frac{n!}{r!(n-r)!} $$

Therefore:

$$ \binom{5}{3} = \frac{5!}{3!(5-3)!} = 10 $$

and

$$ \binom{7}{3} = \frac{7!}{3!(7-3)!} = 35 $$

Please help me understand, thank you.

Answer: $ 10\cdot 35 = 350$.

Answer 1

Check your calculation of ${7 \choose 3}$. You made a typo in the denominator and the result is not $24$, but $35$.

Answer 2

$ {7 \choose 3} $ should be $\frac{7!}{3!\cdot 4!}$ , which is 35.

Answer 3

Note that; $$7C3=\frac{7!}{3!(7-3)!}=35$$

Answer 4

$$\binom53=\frac{5!}{3!(5-3)!}=\frac{5!}{3!2!}=\frac{1\cdot2\cdot3\cdot4\cdot5}{1\cdot2\cdot3\cdot1\cdot2}=\frac{4\cdot5}{1\cdot2}=\frac{20}2=10$$ $$\binom73=\frac{7!}{3!(7-3)!}=\frac{7!}{3!4!}=\frac{7!}{4!3!}=\frac{1\cdot2\cdot3\cdot4\cdot5\cdot6\cdot7}{1\cdot2\cdot3\cdot4\cdot1\cdot2\cdot3}=\frac{5\cdot6\cdot7}{1\cdot2\cdot3}=\frac{5\cdot6\cdot7}6=5\cdot7=35$$ $$\binom53\binom73=10\cdot35=350$$