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My textbook does an interesting cancellation process to simplify the r-combinations.

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How does this process work?

How do you cancel out $4!$ with $19*18*17*16$?

BTW how do you do this 31! with $$?

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It's all about reorganizing terms:

$$\binom{19}{15} = \frac{19!}{15! \cdot (19-15)!}$$

We can decompose the factorial of 19 up to the factorial of 15, i.e, since $19! = 19 \cdot 18 \cdot 17 \cdot 16 \cdot 15 \cdot 14 \cdot 13 \cdot \ldots$ and $15! = 15 \cdot 14 \cdot 13 \cdot \ldots$, we have that $$= \frac{19 \cdot 18 \cdot 17 \cdot 16 \cdot \color{red}{15!}}{\color{red}{15!} \cdot 4!}$$

Simplifying the term in $15!$ and expanding $4!$ we get $$= \frac{19 \cdot \color{red}{18} \cdot 17 \cdot \color{blue}{16}}{\color{blue}{4} \cdot \color{red}{3} \cdot \color{blue}{2} \cdot 1}$$

Since $18$ is a multiple of $3$ and $16$ is a multiple of $4 \cdot 2$ we end up with $$= 19 \cdot 6 \cdot 17 \cdot 2$$

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The $4 \cdot 2$ in the denominator reduces the $16$ in the numerator to $2$ ($\frac{16}{4\cdot 2}=2$). The $3$ in the denominator reduces the $18$ in the numerator to $6$ ($\frac{18}{3}=6$).

It is not clear what you're asking in your "BTW" question, but in general you just look for numbers in the numerator that are divisible by numbers in the denominator and reduce them. If you are calculating a binomial coefficient, then you will always be able to cancel all the numbers in the denominator because binomial coefficients are integers.

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  • $\begingroup$ Okay thank you. re: my btw was how can i cancel 31! in $\frac{31!}{3!}$ for example? $\frac{<s>31!</s>}{3!}$ does not work $\endgroup$ – Evan Kim Apr 18 at 18:05
  • $\begingroup$ Write $31!=31\cdot 30 \cdot 29 \cdots 5 \cdot 4 \cdot 3!$. $\endgroup$ – kccu Apr 18 at 18:28

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