Question

A combination lock has a code consisting of 3 numbers, each of which can be 0 to 39, with numbers repeated. Jillian says that there are only 120 possible codes. Is Jillian correct?

Answer 1

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:

$T = m^{n}$

In this question:

3 trials(values), with 40 options. So

$T = 40^{3} = 64000$

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 2

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