When you encounter probability questions involving dice rolls, you need to systematically count favorable outcomes and divide by total possible outcomes. With two dice rolls, there are $$6 \times 6 = 36$$ total possible outcomes.

To find the probability of getting a sum of at least 10, identify all ways to achieve sums of 10, 11, or 12. For sum = 10: (4,6), (5,5), (6,4) — that's 3 outcomes. For sum = 11: (5,6), (6,5) — that's 2 outcomes. For sum = 12: (6,6) — that's 1 outcome. Total favorable outcomes: 3 + 2 + 1 = 6.

Therefore, the probability is $$\frac{6}{36} = \frac{1}{6}$$, making choice A correct.

Choice B ($$\frac{1}{4}$$) equals $$\frac{9}{36}$$, which would suggest 9 favorable outcomes — this likely comes from incorrectly counting or including impossible combinations. Choice C ($$\frac{5}{18}$$) equals $$\frac{10}{36}$$, suggesting 10 favorable outcomes — this might result from double-counting some combinations or including sums of 9. Choice D ($$\frac{1}{3}$$) equals $$\frac{12}{36}$$, indicating 12 favorable outcomes — this error often occurs when students mistakenly include all sums from 9 to 12 instead of just 10 to 12.

Remember to list outcomes systematically when dealing with "at least" or "at most" probability questions. Draw a table if needed, and always double-check that your favorable outcomes actually satisfy the given condition. The key is being methodical rather than rushing through the counting.

This is a permutation problem where order matters and repetition is not allowed. When you see "order matters" and items are used "at most once," think about arranging distinct objects in sequence.

For the consonants: You need to select and arrange 3 different consonants from 21 available consonants (26 letters minus 5 vowels). Since order matters and no repetition is allowed, you have 21 choices for the first position, 20 for the second, and 19 for the third. This gives $$21 \times 20 \times 19$$, which equals $$\frac{21!}{18!}$$ (since you're taking the first 3 terms of 21 factorial).

For the digits: You need to select and arrange 2 distinct digits from 10 available digits (0-9). With order mattering and no repetition, you have 10 choices for the first digit and 9 for the second. This gives $$10 \times 9$$, which equals $$\frac{10!}{8!}$$.

The total number of passwords is the product: $$\frac{21!}{18!} \times \frac{10!}{8!}$$.

Answer A incorrectly keeps the digits as $$10 \times 9$$ instead of using factorial notation, though mathematically it's equivalent to the correct answer. Answer B uses $$\binom{10}{2}$$ for the digits, which only counts the ways to choose 2 digits but ignores their order. Answer D makes the same mistake as B by using combinations instead of permutations for the digits.

Remember: when order matters and no repetition is allowed, use permutations. The notation $$\frac{n!}{(n-r)!}$$ represents selecting and arranging r objects from n total objects.