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Is there a way to enter the traditional form of a binomial coefficient, i.e. ${n \choose k}$ instead of Binomial[n,k], like I would enter a radical sign $\sqrt{x}$ using Ctrl-Shift-2? I would like to enter it that way in a notebook, and see it in that form in the input as well as output equations, just like the radical. I know that I can use TraditionalForm to convert it on output, but I don't need to do that to see radicals instead of Sqrt.

J. M.'s missing motivation
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Mark Adler
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1 Answers1

18

You could do it using the following (this is the first version of the answer; don't use it if you want the more complete solution below):

SetOptions[EvaluationNotebook[],InputAliases->{"bn"->
 FormBox[TemplateBox[{"\[SelectionPlaceholder]",
  "\[Placeholder]"},"Binomial"],InputForm]}]

Then enter escbnesc to get a placeholder that you can tab through:

pic

Then enter the numbers and press shift-enter to evaluate.

Edit

To make the output appear formatted without invoking TraditionalForm, you'd have to define a wrapper function and its style. I'll set the keyboard shortcut in a way that doesn't erase previous definitions.

ClearAll[myBinomial]

appearance[x_, y_] := TemplateBox[{x, y}, "myBinomial",
  DisplayFunction :> (RowBox[{"(", "\[NoBreak]", 
       GridBox[{{#1}, {#2}}, RowSpacings -> 1, ColumnSpacings -> 1, 
        RowAlignments -> Baseline, ColumnAlignments -> Center], 
       "\[NoBreak]", ")"}] &), 
  InterpretationFunction :> (RowBox[{"myBinomial", "[", 
       RowBox[{#1, ",", #2}], "]"}] &)
  ]

myBinomial[n_?NumericQ, k_?NumericQ] := Binomial[n, k]

myBinomial /: MakeBoxes[myBinomial[n_, k_], StandardForm] := 
 appearance[ToBoxes[n], ToBoxes[k]]

SetOptions[EvaluationNotebook[], 
 InputAliases -> 
  DeleteDuplicates@
   Join[{"bn" -> 
      appearance["\[SelectionPlaceholder]", "\[Placeholder]"]}, 
    InputAliases /. 
      Quiet[Options[EvaluationNotebook[], InputAliases]] /. 
     InputAliases -> {}]
 ]

Now the template appears as before, but when the function arguments aren't numerical, it's also displayed in 2D form. One could add additional definitions to myBinomial depending on what cases you want the 2D display to apply to. Whenever there is no definition for myBinomial, it is left unevaluated and will be displayed in the formatted style.

Update in response to comment:

Similar templates can also be created for other two-dimensional symbols using other types of brackets. As an example, here are shortcuts for StirlingS1 and StirlingS2 that can be entered as escs1esc and escs2esc:

appearanceS1[x_, y_] := 
 TemplateBox[{x, y}, "myStirlingS1", 
  DisplayFunction :> (StyleBox[
      RowBox[{"[", "\[NoBreak]", 
        GridBox[{{#1}, {#2}}, RowSpacings -> 1, ColumnSpacings -> 1, 
         RowAlignments -> Baseline, ColumnAlignments -> Center], 
        "\[NoBreak]", "]"}], SpanMaxSize -> Infinity] &), 
  InterpretationFunction :> (RowBox[{"myStirlingS1", "[", 
       RowBox[{#1, ",", #2}], "]"}] &)]

myStirlingS1[n_?NumericQ, k_?NumericQ] := StirlingS1[n, k]

myStirlingS1 /: MakeBoxes[myStirlingS1[n_, k_], StandardForm] := 
 appearanceS1[ToBoxes[n], ToBoxes[k]]

appearanceS2[x_, y_] := 
 TemplateBox[{x, y}, "myStirlingS2", 
  DisplayFunction :> (StyleBox[
      RowBox[{"{", "\[NoBreak]", 
        GridBox[{{#1}, {#2}}, RowSpacings -> 1, ColumnSpacings -> 1, 
         RowAlignments -> Baseline, ColumnAlignments -> Center], 
        "\[NoBreak]", "}"}], SpanMaxSize -> Infinity] &), 
  InterpretationFunction :> (RowBox[{"myStirlingS2", "[", 
       RowBox[{#1, ",", #2}], "]"}] &)]

myStirlingS2[n_?NumericQ, k_?NumericQ] := StirlingS2[n, k]

myStirlingS2 /: MakeBoxes[myStirlingS2[n_, k_], StandardForm] := 
 appearanceS2[ToBoxes[n], ToBoxes[k]]

SetOptions[EvaluationNotebook[], 
 InputAliases -> 
  DeleteDuplicates@
   Join[{"s1" -> 
      appearanceS1["\[SelectionPlaceholder]", "\[Placeholder]"], 
     "s2" -> appearanceS2["\[SelectionPlaceholder]", 
       "\[Placeholder]"]}, 
    InputAliases /. 
      Quiet[Options[EvaluationNotebook[], InputAliases]] /. 
     InputAliases -> {}]
 ]

In these definitions, I added one additional ingredient: the square and curly brackets for these symbols are not automatically extensible, so they don't grow like the round brackets used in Binomial. To fix this, I added a StyleBox with option SpanMaxSize -> Infinity.

Also, the shortcuts are be appended to the InputAliases without erasing previous definitions. To make the shortcuts work, this step has to be done after defining the wrapper functions appearance and appearanceS1/S2.

With this, you can get output like this:

s1

for StirlingS1 and

s2

for StirlingS2.

Jens
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  • Awesome! Though it comes out as Binomial[n,k] if it's not numbers. Can I get it to display in the traditional form without having to use TraditionalForm? – Mark Adler Apr 07 '14 at 05:30
  • @MarkAdler Yes you can do that, too, but I wouldn't recommend it. To make the output behave differently, one would have to do something similar to what I did in this answer, but for StandardForm. It would require defining a wrapper function myBinomial so you don't have to modify the protected Binomial itself. Then the Format for that wrapper function can be defined to be what you want. I'll leave it up to you to adapt that other answer, since this question was only about input. – Jens Apr 07 '14 at 05:55
  • I'll look at that. Thanks. By the way, this question was also about the output. You'd have to read the entire body of the question. The title of the question is abbreviated and necessarily incomplete. – Mark Adler Apr 07 '14 at 06:06
  • @MarkAdler Ah, I thought you'd be OK with a display that only works in TraditionalForm. I added something to make it work in StandardForm. – Jens Apr 07 '14 at 14:47
  • Awesome squared. Works like a charm. Thank you. – Mark Adler Apr 07 '14 at 22:32
  • Nice. I suppose this can be modified to allow a user to enter Stirling numbers in Karamata's notation (i.e. $\left[n\atop k\right]$ for the cycle numbers, and $\left{n\atop k\right}$ for the subset numbers), right? – J. M.'s missing motivation Jun 08 '15 at 06:46
  • @Guesswhoitis. I think so - should I try to add that? I'll have time later today... – Jens Jun 08 '15 at 18:24
  • Only if it's convenient for you. :) It'd definitely be nice. – J. M.'s missing motivation Jun 08 '15 at 18:48
  • @Guesswhoitis. I added your suggested notation for the Stirling numbers. – Jens Jun 09 '15 at 20:32
  • My only disappointment is that I cannot give you another upvote. This looks wonderful! :) (The cycle numbers are actually (-1)^(n - k) StirlingS1[n, k], but that is a minor nit in an otherwise good answer.) – J. M.'s missing motivation Jun 09 '15 at 22:39