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From: poslfit@gmail.com (John J. Chew III)
Date: Sun, 19 Oct 1997 00:25:38 -0400
Subject: puzzle solution, bahuvrihis, probability
Message-id: <199710190425.AAA22007@coxeter.math.toronto.edu>

The answer to my recently posted puzzle can be found at:

http://www.math.utoronto.ca/~jjchew/scrabble/lists/unistems62.html

Sam Kantimathi recently suggested that a false compound might be called a BAHUVRIHI#. I feel compelled to point out that a bahuvrihi is one of a pentad of grammatical terms describing compound word formation, coined by Sanskrit pedants a few thousand years ago and still in use today by English-speaking linguists and computer scientists. A bahuvrihi is not a false compound, but rather a possessive one, in which the composition of two words NUMB+SKULL denotes not an instance of the latter (that is, not a kind of skull), but rather one who possesses such an item. Other types of compounds are karmadharayas* (descriptive compounds, such as BLACK+BIRD), tatpurushas# (dependent compounds, such as BAND+WIDTH, NIT+PICKER), and dvandvas# (copulative compounds, such as PRINCE-CONSORT). I believe the fifth group are the prepositional compounds (like COMPOUND itself), but haven't been able to find what the Sanskrit term is. Sam, do you know?

(I do of course find it appalling that people in this continent have put up with OSPD for so long, when it is missing so many of these everyday words.)

Finally, to reply to a private query, with an answer that might be of remotely general interest (but mainly for the archive):

How do you calculate the probability of drawing a word, say TZITZIT, out of the bag, taking into account that you may use blanks for some of the letters?

In a standard English bag, there are (100*99*98*97*96*95*94)/(7*6*5*4*3*2*1) = 16,007,560,800 ways of choosing any seven tiles out of the bag.

Since there are nine Is, six Ts and one Z, there are zero ways of drawing TZITZIT without using any blanks. That's (9*8)/(2*1) = 36 ways of choosing the Is, (6*5*4)/(3*2*1) = 20 ways of choosing the Ts, and 0 ways of choosing the Zs: 36*20*0 = 0.

How can we do it with one blank? The blank has to be a Z. 36 choices for the Is, 20 choices for the Ts, 1 choice for the Z, 2 choices for the blank. 36*20*1*2 = 1440 ways all told.

How about with both blanks? We could use blanks for both Zs (36 choices for the Is, 20 choices for the Ts, 1 choice for the blanks for a total of 36*20*1 = 720), one blank for a Z and one blank for an I (9 for the Is, 20 for the Ts, 1 for the Z, 1 for the blanks, 9*20*1*1 = 180), or one blank for a Z and the other for a T (36 for the Is, (6*5)/(2*1)=15 for the Ts, 1 for the Z, 1 for the blanks, 36*15 = 540).

Grand total: 0+1440+720+180+540 = 2880

Probability = 2,880/16,007,560,800 or about one in 5.5 million.

John

John Chew (poslfit on MD/WD/PD)
poslfit@gmail.com * http://www.math.utoronto.ca/~jjchew