Respuesta :
Answer:
450,000 codes
Step-by-step explanation:
Number of digits in the access code = 6
Conditions:
- First digit cannot be 2
- Last digit must be odd
Since, there are 10 digits in total from 0 to 9 and first digit cannot be 2, there are 9 ways to fill the place of first digit.
Last digit must be odd. As there are 5 odd digits from 0 to 9, there are 5 ways to fill the last digit.
The central 4 digits can be filled by any of the 10 numbers. So, each of them can be filled in 10 ways.
According to the fundamental rule of counting, the total possible codes would be the product of all the possibilities of individual digits.
Therefore,
Number of possible codes = 9 x 10 x 10 x 10 x 10 x 5 = 450,000 codes
Hence, 450,000 different codes are possible for the lock box
Locks with digits can usually have its digits valued from 0 to 9 (10 choices).
The count of different codes for the considered lock with given constraints is 450,000
What is the rule of product in combinatorics?
If a work A can be done in p ways, and another work B can be done in q ways, then both A and B can be done in [tex]p \times q[/tex] ways.
Remember that this count doesn't differentiate between order of doing A first or B first then doing other work after the first work.
Thus, doing A then B is considered same as doing B then A
For given case, we have:
First digit cannot be 2, so it can be from {0,1,3,4,5,5,6,7,8,9} = 9 choices
Second to fourth digits have 10 choices each.
Last digit(sixth) needs to be odd so it can be chosen from {1,3,5,7,9} = 5 choices.
Thus, by using rule of product, we get:
Total possible different codes = [tex]9 \times 10 \times 10 \times 10 \times 10\times 5 = 450,000[/tex]
Thus,
The count of different codes for the considered lock with given constraints is 450,000
Learn more about rule of product here:
https://brainly.com/question/2763785
