Answer:
There are 64,000 possible codes. Jillian is incorrect, since he added the possible options in each trial, instead of multiplied them(he did 40+40+40 instead of 40*40*40).
Step-by-step explanation:
From 0 to 39, there are 40 numbers.
The code has three values:
V1 - V2 - V3
When we have n trials with m possible options, the total number of opitons is:
[tex]T = m^{n}[/tex]
In this question:
3 trials(values), with 40 options. So
[tex]T = 40^{3} = 64000[/tex]
There are 64,000 possible codes. Jillian is incorrect, since he added the possible options in each trial, instead of multiplied them(he did 40+40+40 instead of 40*40*40).
Answer:
Step-by-step explanation:
The lock has a code that consists of 3 number
It is given that the number in the lock code can be used between 0 to 39
Thus, out of total 40 set of numbers (i.e 0-39), the numbers can be repeated.
This means for all three code numbers the opportunity of choosing a number is same i.e. between 0-39
Now, the first digit of the code can be any number between 0-39
Like wise the second and third digit of the code can be any number between 0-39
Thus. the possible number of codes with repetition allowed are
40 * 40* 40 = 64000
Hence, Jillian is not correct
Jillian has summed up the possible number of codes for each digit instead of multiplying it
