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4 votes
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? If not, explain her error.

User Cortright
by
8.0k points

2 Answers

4 votes

Answer:

No, Jillian is not correct. According to the fundamental counting principle, the total number of possible codes would be (40)(40)(40) = 64,000. Jillian multiplied by 3 instead of multiplying the number of possibilities for each digit.

Explanation:

User Kikerrobles
by
7.3k points
6 votes

Answer:

Jillian is not correct

Jillian has summed up the possible number of codes for each digit instead of multiplying it

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

User Apgsn
by
8.4k points
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