a) Consider the 7-digit registration code to be an arrangement of 7 cells to be filled using the given digits.
In the first cell, one can write any of the digits; on the other hand, there are only 6 digits available to fill the second cell (no number can be used more than once). Therefore, there are 5 digits that can be used in the third cell and so on; thus, there is a total of
[tex]7*6*5*4*3*2*1=7!=5040[/tex]
5040 different registration codes.
b) The 5040 different combinations found above are equally probable.
There are only 3 available even numbers (2, 4, and 6); therefore, we need to find the number of combinations such that none of the first three digits is equal to 2, 4, or, 6.
Thus, using a diagram,
There are 4 possible numbers that one can fit in the first cell (1,5,7, or 9), in the second cell, one can fit 3 numbers (any of the remaining ones from cell 1), and so on.
In the fourth cell (first cell in blue), one can fit any even number plus a remaining odd number from cell 3.
Therefore, the total number of codes such that their first three digits are not even are
[tex]4*3*2*4*3*2*1=576[/tex]
Then, the corresponding probability is
[tex]P=\frac{576}{5040}=\frac{4}{35}[/tex]
The answer to part b) is 4/35
