Question

a safe has a 4-digit lock code that does not include zero as a digit and no digit is repeated. what is the probability that the lock code consists of all even digits?
to find the total number of outcomes for this event, find the permutation of \boxed{} things taken 4 at a time.
the total number of outcomes is \boxed{}.
the total number of favorable outcomes is a permutation of \boxed{} things taken 4 at a time.
the probability that the lock code consists of all even digits is \boxed{} out of 3,024.

Explanation:

Step1: Determine total digits available

Digits not including zero: 1 - 9, so 9 digits.

Step2: Calculate total permutations (n=9, r=4)

Permutation formula: \( _nP_r=\frac{n!}{(n - r)!} \)
\( _9P_4=\frac{9!}{(9 - 4)!}=\frac{9!}{5!}=9\times8\times7\times6 = 3024 \)

Step3: Determine even digits available (no zero)

Even digits from 1 - 9: 2, 4, 6, 8 → 4 digits.

Step4: Calculate favorable permutations (n=4, r=4)

\( _4P_4=\frac{4!}{(4 - 4)!}=\frac{4!}{0!}=4\times3\times2\times1 = 24 \)

Step5: Find probability

Probability = favorable / total = \( \frac{24}{3024} \) (but we just need the numerator for the last blank as total is 3024)

Answer:

s (filling the blanks in order):
First blank: 9
Second blank: 3024
Third blank: 4
Fourth blank: 24