Combinatorics arrangement question

Question

Taken the word to be "Logarithm", how many ways can this word be arranged alphabetically? For example "Ail" is valid but "mhi" is not. I know how to find how many words can be arranged using e*n!, but have no idea how to find out how many words alphabetically. Many thanks

Answer 1

The word "Logarithm" is made of nine different characters. If we select $k$ characters out of these nine we have only one way to arrange them in alphabetical order, hence the answer is given by: $$ \sum_{k=0}^{9}\binom{9}{k} = 2^9 = \color{red}{512} $$ where we accounted also for the empty word.

Answer 2

$l=5,o=7,g=2,a=1,r=8,i=4,t=9,h=3,m=6$

$123456789$ from this number

In how many ways can I remove $1$ digit? ${9 \choose 1}$

In how many ways can I remove $2$ digits? ${9 \choose 2}$

In how many ways can I remove $3$ digits? ${9 \choose 3}$

...

$9+{9 \choose 2}+{9 \choose 3}+{9 \choose 4}+{9 \choose 5}+{9 \choose 6}+{9 \choose 7}+{9 \choose 8}=9+36+84+126+126+84+36+9=510$

So there are $510$ ways of arranging the letters (where the number of letters is less than $9) +1$ (when all the letters are considered)=$511$ ways