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I am trying to get used to the binomial notation. The general forumla for it is:

$$\binom{k}{n}=\dfrac{k(k-1)(k-2)\cdots(k-n+1)}{k!}$$

So let take $\binom{k-4}{2}$, that is going to equal to \begin{align} & \frac{k(k-1)(k-2)\cdots(k-4-2+1)}{2!} \\[8pt] = {} &\frac{k(k-1)(k-2)\cdots(k-5)}{2} \end{align}

Is this the correct result?

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    $\begingroup$ No. You have a $k-4$ on the top level of your binomial coefficient. Thus, the product in the numerator should start at $k-4$, not at $k$. $\endgroup$ – darij grinberg Nov 15 at 22:29
  • $\begingroup$ $$\binom{k-4}2=\frac{(k-4)\cdot(k-5)}2.$$ $\endgroup$ – Maximilian Janisch Nov 15 at 22:31
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    $\begingroup$ The denominator should be $n!,$ not $k!$ in your formula for $\binom k n.$ $\endgroup$ – Thomas Andrews Nov 15 at 22:32
  • $\begingroup$ Strictly speaking, the choice of letters $n$ and $k$ has no intrinsic meaning. But if they both occur in a binomial coefficient like this, you will almost certainly see $n$ at the top and $k$ at the bottom: $n$ will be the total number of objects to choose from, and $k$ will be the number of objects that you want to choose. This is just notational convention, but it is as well to get used to it from the start. In fact, your mistaken denominator ($k!$ for $n!$) is probably a direct result of this notational confusion. $\endgroup$ – TonyK Nov 16 at 11:34
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We have that

$$\binom{k}{n}=\dfrac{k!}{n!(k-n)!}=\dfrac{k(k-1)(k-2)...(k-n+1)}{n!}$$

and then

$$\binom{k-4}{2}=\dfrac{(k-4)!}{2!(k-6)!}=\dfrac{(k-4)(k-5)(k-6)!}{2!(k-6)!}=\dfrac{(k-4)(k-5)}{2}$$

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  • $\begingroup$ is $(k-4)!=(k-1)(k-2)(k-3)(k-4)$ and $(k-6)!=(k-1)(k-2)(k-3)(k-4)(k-5)(k-6)$? $\endgroup$ – James Warthington Nov 15 at 22:55
  • $\begingroup$ We have that $(k-4)!=(k-4)(k-5)(k-6)(k-7)...3\cdot 2 \cdot 1=(k-4)(k-5)(k-6)!$ $\endgroup$ – user Nov 15 at 22:59
  • $\begingroup$ @JamesWarthington, no. $n!$ is the product of all positive integers up to, and including, $n$. It could have $2$ terms or $1,000$. If $k=100$ then $(k-4)!$ is the product of the integers $1$ through $96$. $\endgroup$ – John Nov 15 at 23:01
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    $\begingroup$ It's read "$k-4$ factorial equals $k-4$, times $k-5$, times $k-6$ factorial." Let's take $k=100$. Then we have "The product of the first $96$ integers equals $96$, times $95$, times the remaining ninety-four integers multiplied together." The whole purpose of writing it that way is to cancel $(k-6)!$ top and bottom. $\endgroup$ – John Nov 15 at 23:06
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    $\begingroup$ @JamesWarthington Usually when top number is less than the bottom we take by definition binomial coefficient equal to zero. $\endgroup$ – user Nov 15 at 23:23
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Compare

$$\binom{j}{2} = \frac{j(j-1)}{2}$$

and substitute $j=k-4$.

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The binomial coefficient $\binom{a} {n} $ is defined for all $a\in\mathbb {C} $ and all $n\in \mathbb {N} $ via $$\binom {a} {n} =\frac{a(a-1)(a-2)(a-3)\dots (a-(n-1))}{n!}$$ The numerator contains $n$ factors starting with $a$ and decreasing by $1$ as one moves from one factor to the next. The last factor thus is $a-(n-1)=a-n+1$. The denominator also follows same pattern with the first factor being $n$.

Also by convention we define $\binom{a} {0}=1$. So for $\binom{k-4}{2}$ your numerator should start start with $k-4$ and the next factor is $k-5$ and your stop because only two factors are needed. The denominator is $2\cdot 1$ and thus we have $$\binom{k-4}{2}=\frac{(k-4)(k-5)}{2}$$

When $a$ is also a positive integer then you can prove that $$\binom{a} {n} =\frac{a!} {n! (a-n)!} $$

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Remember also that in the expansion of the binomial, there are as many factors upstairs as downstairs, as long as you include the factors equal to $1$.

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