I often say to my students: There is a simple way to find binomial formula ,and the way is :$$ \begin{pmatrix} n \\ k \end{pmatrix} =\dfrac{\underbrace{n.(n-1).(n-2)...}_{\text{k-terms decreasing}}}{\underbrace{1.2.3...k}_{\text{k-terms increasing}}}$$ for example: $$\begin{pmatrix} 5 \\ 3 \end{pmatrix}= \frac{5!}{3!(5-3)!} \to \\ \begin{pmatrix} 5 \\ 3 \end{pmatrix}= \frac{5.4.3}{1.2.3}=5.2=10 $$ for example $$\begin{pmatrix} n \\ 2 \end{pmatrix}= \frac{n(n-1)}{1.2} =\frac{n(n-1)}{2}\\\begin{pmatrix} n \\ 3 \end{pmatrix}= \frac{\underbrace{n(n-1)(n-2)}_{\text{3-terms decreasing}}}{\underbrace{1.2.3}_{\text{3-terms increasing}}} $$ or $$\begin{pmatrix} 11 \\ 7 \end{pmatrix}\underbrace{=}_{\text{w.r.t. to }\begin{pmatrix} n \\ k \end{pmatrix}=\begin{pmatrix} n \\ n-k \end{pmatrix}} \begin{pmatrix} 11 \\ 4 \end{pmatrix}=\frac{}{}$$ I tel this for speedy work with binomial to k-11,k-12 students.Is this a good way to teach binomial calculation for the students ? (instead of formal calculation $\frac{n!}{r!(n-r)!} $) .If not tell my why ?

I am thankful for your ideas , opinion ,...in advance.


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    $\begingroup$ i think you should move this to matheducators.stackexchange.com $\endgroup$ – lasen H Aug 3 '17 at 11:28
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    $\begingroup$ What's your actual question? $\endgroup$ – Lord Shark the Unknown Aug 3 '17 at 11:29
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    $\begingroup$ Basic definition should be used ..but tricks can be used and taught for speed. $\endgroup$ – Pranita Gupta Aug 3 '17 at 11:36
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    $\begingroup$ The falling factorial definition makes more sense for the generalized binomial theorem, when you have things like: $$\sqrt{x+y}=\sum_{k=0}^\infty\binom{1/2}kx^{\frac12-k}y^k$$ $\endgroup$ – Simply Beautiful Art Aug 3 '17 at 11:58
  • $\begingroup$ This is the method I personally use to calculate binomial coefficients by hand or mentally. It is comparable to generating Pascal's triangle in terms of ease for not-too-big numbers. For even bigger numbers, there is an nCr option in good calculators. $\endgroup$ – anon Aug 3 '17 at 15:30

I often say to my students: There is a simple way to find binomial formula ,and the way is :$$ \binom n k =\dfrac{\overbrace{n\cdot (n-1)\cdot (n-2)\cdots}^{\text{k-terms decreasing}}}{\underbrace{1\cdot 2\cdot 3\cdots k}_{\text{k-terms increasing}}}$$

Sure, but to be clearer try it like this:

$$ \dbinom n k ~=~\dfrac{n^{\underline k}}{k^{\underline k}}~=~\dfrac{\overbrace{n\cdot (n-1)\cdot (n-2)\cdots (n-k+1)}^{\text{k-terms decreasing}}}{\underbrace{k\cdot (k-1)\cdot (k-2)\cdots 1}_{\text{k-terms decreasing}}\qquad\qquad}$$

Where $n^{\underline k}$ is called the $k$ falling-factorial of $n$.   This is excellent if you want to calculate by hand, and will give the students a practical feel for the numbers.

However, modern personal computers (or phones even) can easily handle most factorials that you'll want to use, so using the factorial function will be quicker (that is, unless your package has access to a binom[,] function which probably uses the falling factorial to calculate it anyway).

However, there is a reason to have the students think of the binomial coefficient as $\frac{n!}{k!(n-k)!}$.   This is because when used in formulas that multiply or divide binomial coefficients expressed like so, there is often a lot of convenient cancelling.

Still knowing the falling factorial expression will prove useful if you cover generalised binomial expressions.

$$\dbinom{1/2}{5}=\dfrac{0.5\cdot (-0.5)\cdot(-1.5)\cdot(-2.5)\cdot(-3.5)}{5!}=\dfrac 7{256}$$

And such.

So knowing the falling-factorial form cannot hurt, and can help, but should be taught in addition to rather instead of the factorial form.

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    $\begingroup$ @Khosrotash: Graham's answer is very good. I think you can parallel it with your method by introducing (soon or later) that $k! = k^{\,\underline {\,k\,} } = 1^{\,\overline {\,k\,} } $ and with the classical definition by noting that $n^{\,\underline {\,k\,} } = n!/\left( {n - k} \right)!$ $\endgroup$ – G Cab Aug 3 '17 at 15:00

For small $n,k$ such that the factorials are in range, the standard formula is appropriate.

For larger $n$ the falling factorial approach is indeed more efficient (and computing for $k$ or $n-k$ should be considered).

When you need the coefficients for successive $k$ and $n$, the recursive method (leading to the construction of Pascal's triangle) is better.


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