124452

== SLUMGULLION ==

Combinatorics, arrangements (edited)

"How many ways can the letters in the word SLUMGULLION be arranged so that the three L’s precede all the other consonants?"

Here is an automaton that will generate the required words. A good word has three parts : A is before the third L, then the third L, then the suffix B that contains no L's.

the equations are :

$$A = 1 + (l+i+o+u).A$$

$$B = 1 + (g + m + n + s + i + o + u) .B$$

$$W = A.l.B$$

the generating function for W is

$${1 \over 1- (l+i+o+u)} .l. {1 \over 1- (g+m+n+s + i + o + u)}$$

and we are interested in the coefficient of

$$l^3.g.m.n.s.u^2.i.o$$ which is $$95040$$.


 * The third L** is necessary and it may be find (hidden) both in solution and in algorithm.

As one may see, such a problem involves only the sum and the product rules, and some coefficient stuff.

But, since 11 = (1+1+2) + (3+1+1+1+1) and there are **four types of letters** I can not imagine a shorter answer than the first, which is a product of four (factorial) factors.

U-type : 2 letters I-type : 2 letters S-type : 4 letters, consonants L type : 3 letter, a multiple L

we have:

$$11= 2+2+3+4, $$

$$7 = 3+4, $$

$$4 = 4, $$

$$2 = 2$$