Six number of fair dice are rolled simultaneously. What is the probability that each face shows up with a different number?

Correct option is C

Correct Option: C. $$\frac{6!}{6^6}$$​​
Explanation:
The problem requires finding the probability that in a single throw of six dice, no two dice show the same number. We calculate this by dividing the number of favorable permutations by the total number of possible outcomes.
Information Booster:
1. Total Outcomes: Each of the 6 dice can land in 6 different ways independently. Thus, the total sample space is $$6 \times 6 \times 6 \times 6 \times 6 \times 6 = 6^6$$​.
2. Favorable Outcomes: To have all different numbers, we must arrange the 6 unique faces (1, 2, 3, 4, 5, 6) across the 6 dice. This is a permutation of 6 items, calculated as $$6 \times 5 \times 4 \times 3 \times 2 \times 1 = 6!$$​.
3. Calculation: The probability is $$\frac{\text{Favorable Outcomes}}{\text{Total Outcomes}} = \frac{6!}{6^6}.$$​​
Additional Knowledge:
1. Numerical Value: $$\frac{720}{46,656} \approx 0.0154 (or 1.54\%$$​).
2. Pigeonhole Principle: If we had 7 dice, it would be impossible to have all different numbers because there are only 6 possible faces (probability would be 0).
3. Comparison: If the question asked for the probability that all dice show the same number (e.g., all 1s, all 2s), the numerator would be 6 (since there are 6 such outcomes), resulting in $$\frac{6}{6^6} = \frac{1}{6^5}$$​.