AHL 1.10—Perms and combs, binomial with negative and fractional indices

There are 10 digits (0–9) and 26 uppercase letters (A–Z), giving $10+26=36$ possible characters for each position.

Since repetition is allowed and order matters, each of the 5 positions independently has 36 choices. The total number of codes is $36\times 36\times 36\times 36\times 36=36^5$.

$36^5=36\times 36\times 36\times 36\times 36=60466176$

The number of distinct PIN codes is $36^5=60466176$.

We need the total number of ordered 5-character strings from an alphabet of 36 symbols with repetition allowed.

This simply multiplies the alphabet size by the code length, which is not a valid counting rule. Eliminated.

${}^{36}{P}_5$ counts ordered arrangements without repetition. Since repetition is allowed here, this formula undercounts. Eliminated.

$\left(\genfrac{}{}{0ex}{}{36}{5}\right)$ counts unordered selections without repetition — both assumptions are wrong for a PIN code. Eliminated.

With repetition allowed and order mattering, each position has 36 independent choices, giving $36^5=60466176$. Correct.

We are arranging all 5 distinct trophies in a row. Order matters (different positions give different arrangements).

The number of ways to arrange $n$ distinct objects is $n!$. Here $n=5$, so we compute $5!$.

$5!=5\times 4\times 3\times 2\times 1=120$

There are $120$ possible arrangements of the 5 trophies.

We want the number of ordered arrangements of all 5 distinct objects.

${}^5{P}_2$ only arranges 2 of the 5 objects, not all 5. Eliminated.

$\left(\genfrac{}{}{0ex}{}{5}{2}\right)$ chooses 2 objects without regard to order — completely wrong context. Eliminated.

$5^2$ would apply if we had 2 positions each with 5 choices (with repetition), which is not the case here. Eliminated.

Arranging all 5 distinct objects gives $5!=120$. Correct.