We can make 55 more codes in Method B than in Method A. In Method A we can make 45 codes and in Method B we can make 100 codes.
Method A:
For the first odd digit, we can have 1,3,5,7,9
For the last two digit we have options 11,22,33,44,55,66,77,88,99
So, for making the code we have 5 options for the first digit and 9 options for the second and the third digit.
Therefore, if we choose 1 as the first digit, we have 9 options for the second and third digits. Similarly for 3,5,7,9
Hence we will have 45 options in Method A
Method B:
For the first digit, we can have 1,3,5,7,9 as options.
For the second digit, we have 2,4,6,8 as an option
For the third digit, we have 1,3,5,7,9 as an option
Therefore, for the first digit, we have 5 options, the second digit 4 options and 5 options for the third digit.
So, we have 5*4*5 options
Hence, we have 100 options available in Method B
Learn more about permutations:
brainly.com/question/14368656
#SPJ9
