Answer: There are 6561 posible combinations.
Step-by-step explanation:
the lock has four dials.
And we know that the first dial has 9 posible numbers (0, 1, ... , 9)
the second dial also has 9 posible numbers, and it is the same for the other two.
The total number of combinations is the product of the posibilities for each dial:
c = 9*9*9*9 = 9^4 = 6561
You can use rule of product from combinatorics for counting the number of possible combinations for given lock.
The count of possible combinations for given lock is
10,000
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 context)
There are total 10 digits (0 to 9), available for each dial.
You can use the fact that for each digit of first dial, there can be 10 digit of second dial (thus 10 times 10 = 100 ways) and for each pair of digits for first and second dials, there can be 10 digits of third dial(100 times 10 = 1000 ways) and for each 3 sized pair of those 3 dials, there are 10 digits of fourth dial(1000 times 10 = 10000 ways).
Thus, total combinations will be
[tex]10 \times 10 \times 10 \times 10 = 10,000[/tex]
Or using rule of product, as there are 10 digits for each dial, thus, [tex]10 \times 10 \times 10 \times 10 = 10,000[/tex] ways.
Thus,
The count of possible combinations for given lock is
10,000
Learn more about rule of product of combinatorics here:
https://brainly.com/question/2763785
