The pin a cellphone consists of 6 distinct digits (no repetitions)selected from 0-9. How many possible combinations are there?
To find the number of possible combinations, we can use the concept of permutations since order matters and no digit can be repeated. The formula for permutations is:
P(n, r) = n! / (n - r)!
Where:
n = total number of digits to choose from
r = number of digits to select
In this case, we have 10 digits to choose from (0-9) and we need to select 6 digits without repetition.
Plugging the values into the formula:
P(10, 6) = 10! / (10 - 6)!
Now, let's calculate this:
10! = 10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1 = 3,628,800
(10 - 6)! = 4! = 4 x 3 x 2 x 1 = 24
Substituting the values back into the formula:
P(10, 6) = 3,628,800 / 24 = 151,200
Therefore, there are 151,200 possible combinations for a cellphone PIN with 6 distinct digits selected from 0-9.