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

My work is below: Can someone also solve this ONLY using the multiplication rule, permutations, and permutations with repetitions?

We have 3 L's and the other 4 consonants are S,M,G,N. That is, our consonants are LLLSMGN, call them all X for the moment. Then we have XXXXXXXUUIO. The number of arrangements of these letters is $\frac{11!}{7!2!}$. Hence the answer is $4!*\frac{11!}{7!2!}$ since there are $4!$ ways to arrange the 4 consonants other than the L's.

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    Your reasoning is correct. I also solved the problem a di$f$ferent way (using combinations) and got the same answer.2012-03-26

2 Answers 2

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Your argument is fine (as previously noted), and there's probably no significantly better approach to counting them.

If you want to double-check your result, here's some GAP code which can list all the possibilities.

A:=["S","L","U","M","G","U","L","L","I","O","N"]; T:=Arrangements(A,Size(A)); count:=0;  for P in T do    # where the last L is   a:=Maximum(Positions(P,"L"));    # where the first non-L consonant is   b:=Minimum(Position(P,"S"),Position(P,"M"),Position(P,"G"),Position(P,"N"));    if(b>a) then     count:=count+1;     Print(P,"\n");   fi;  od;  Print(count,"\n"); 

and it found 95040, matching your result $4! \frac{11!}{7!\ 2!}$.

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skematon

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, $

$2 = 2$.