# No. of possible 'words'

• January 15th 2012, 05:15 AM
Punch
No. of possible 'words'
In this question, a 'word' is defined to be any set of letters in a row, whether of not it makes sense. Find how many different 'words' can be made using only 5 letters of the word SYLLABUS

I tried to separate them into 4 cases,

1) all different letters
2) 2 'S'
3) 2 'L'
4)2 'S' and 2 'L'
• January 15th 2012, 12:14 PM
Soroban
Re: No. of possible 'words'
Hello, Punch!

Quote:

In this question, a 'word' is defined to be any set of letters in a row,
whether of not it makes sense. . Find how many different 'words'
can be made using only 5 letters of the word SYLLABUS.

I tried to separate them into 4 cases,

1) all different letters
2) two S's
3) two L's
4) two S's and two L's

An excellent game plan! How far did you get?

We have 8 letters, 6 distinct: . $A,B,LL,SS,U,Y$

1) Five different letters
. . Choose 5 of the 6 letters and permutate them:. $_6P_5 \,=\,720$

2) Two S's: . $S\,S\,\_\,\_\,\_$
. . Choose 3 of the other 5 letters:. $_5C_3 \,=\,10$
. . and permute the 5 letters:. $10\cdot\tfrac{5!}{2!} \,=\,600$

3) Two L's:. $L\,L\,\_\,\_\,\_$
. . Choose 3 of the other 5 letters:. $_5C_3\,=\,10$
. . and permutate the 5 letters:. $10\cdot\tfrac{5!}{2!}\,=\,600$

4) Two S's, two L's:. $S\,S\,L\,L\,\_$
. . Choose 1 of the other 4 letters:. $_4C_1\,=\,4$
. . and permutate the 5 letters:. $4\cdot\tfrac{5!}{2!\,2!}\,=\,120$

There are:. $720 +600 + 600 +120 \:=\:2040$ "words".