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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

 Jan 2, 2021
 #1
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+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.

 Jan 2, 2021
 #2
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+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.

 Jan 2, 2021

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