In how many ways can one write the numbers 1, 2, 3, 4, 5, and 6 in a row so that given any number in the row, all of its divisors (not including itself) appear to its left?

plz help i am stuck

Guest Jan 2, 2021

#1**+1 **

I wrote a computer program:

For a = permutation[1,2,3,4,5,6]

good = 1;

if i > j and a[j] % a[i] = 0 then good = 0;

count = count + good

write(output,count);

output = 7.

Guest Jan 2, 2021

#2**+1 **

1,2,3,4,5,6

1 has to go first.

1, 23456

2,3, and 5 are prime so they can go in any order in relation to one another.

4 has to go after 2

6 has to go after both 2 and 3

Each time there are 5 spots for the 5

1,2,3,4,6 5ways with the 5 included

1,2,3,6,4 5ways with the 5 included

1,2,4,3,6 5ways with the 5 included

1,3,2,4,6 5ways with the 5 included

1,3,2,6,4 5ways with the 5 included

So I get 25 ways.

Melody Jan 2, 2021