SOLUTION: Freddie has forgotten the 6-digit code that he uses to lock his briefcase. He knows that he did not repeat any digit and that he did not start his code with a zero. (i) Find the

Algebra ->  Permutations -> SOLUTION: Freddie has forgotten the 6-digit code that he uses to lock his briefcase. He knows that he did not repeat any digit and that he did not start his code with a zero. (i) Find the       Log On


   



Question 1205223: Freddie has forgotten the 6-digit code that he uses to lock his briefcase. He knows that he did not
repeat any digit and that he did not start his code with a zero.
(i) Find the number of different 6-digit numbers he could have chosen.
Freddie also remembers that his 6-digit code is divisible by 5.
(ii) Find the number of different 6-digit numbers he could have chosen.

Answer by ikleyn(52780) About Me  (Show Source):
You can put this solution on YOUR website!
.
Freddie has forgotten the 6-digit code that he uses to lock his briefcase. He knows that he did not
repeat any digit and that he did not start his code with a zero.
(i) Find the number of different 6-digit numbers he could have chosen.
Freddie also remembers that his 6-digit code is divisible by 5.
(ii) Find the number of different 6-digit numbers he could have chosen.
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       The solution and the answer to question (i)



The number of different 6-digit codes he could have chosen is

     9*9*8*7*6*5 = 136080.



Any of 9 = 10-1 digits can be in 1st position (any except of 0).

Any of 9 remaining digits can be in 2nd position (0 is allowed now in the 2nd position).

Any of 8 remaining digits can be in 3rd position.


And so on, till the end.



       The solution and the answer to question (ii)



The number of different 6-digit codes he could have chosen is

     9*8*7*6*5*1 + 8*8*7*6*5*1 = 15120 + 13440 = 28560.


First addend is for codes that end with 0 - for them, there is only 1 choice for the 6th digit.

Second addend is for codes that end with 5 - so, there is only 1 choice for the 6th digit.

Solved.