The Multiplication Principle

Hold one shirt still. It pairs with each of the $4$ pairs of pants, so that single shirt gives $4$ outfits. Every one of the $5$ shirts gives its own $4$, and none of those repeat, so the total is $5\times 4=\boxed{{\displaystyle 20}}$. Equal-sized groups get multiplied, which beats adding $4+4+4+4+4$ by hand.

Four menus means four slots, holding $2$, $3$, $4$, and $6$ choices. Multiply straight across. $2\times 3=6$, then $6\times 4=24$, then $24\times 6=144$, so you can build $\boxed{{\displaystyle 144}}$ heroes. Picking a body type does not remove any hair or outfit options, so every stage keeps its full count and the slots multiply.

Each of the $3$ dials can show any of the $6$ symbols, so multiply $6\times 6\times 6=6^3=216$. The lock has $\boxed{{\displaystyle 216}}$ codes. Repeats being allowed is what keeps the count at $6$ every time, since using a symbol on one dial does not use it up.

Four slots, letter, letter, digit, digit. A letter slot has $26$ choices and a digit slot has $10$, and repeats are allowed so no slot loses anything to the ones before it. The two letters give $26\times 26=676$, the two digits give $10\times 10=100$, and $676\times 100=67600$. There are $\boxed{{\displaystyle 67600}}$ plates.

Fill $6$ shelf slots left to right. The first slot has $6$ books to choose from, then $5$ are left, then $4$, $3$, $2$, and $1$. Multiply, $$6\times 5\times 4\times 3\times 2\times 1=720,$$ so there are $\boxed{{\displaystyle 720}}$ orders. Placing a book uses it up, so each slot has one fewer choice than the last. That countdown product is $6!$.

Any of the $9$ members could be president. That leaves $8$ people for vice president and $7$ for treasurer, so multiply $9\times 8\times 7$. Since $9\times 8=72$ and $72\times 7=504$, the jobs can be filled in $\boxed{{\displaystyle 504}}$ ways. Three jobs means three factors, so stop the countdown after three instead of running it down to $1$.

Count the ones you do not want. The $3$-digit numbers with all three digits the same are $111,222,333,444,555,666,777,888,999$, which is $9$ of them. Take those from the $900$ and you get $900-9=\boxed{{\displaystyle 891}}$. Every three-digit number is either all-same or not, so subtracting the small group leaves exactly the big one. That is the total-minus-bad move from $10.1$.

Three boxes, hundreds, tens, units. The hundreds box cannot be $0$, so it has the $9$ choices $1$ through $9$, and the tens and units boxes each take any of $10$ digits. Multiply, $$9\times 10\times 10=900,$$ so there are $\boxed{{\displaystyle 900}}$ three-digit numbers. Total minus bad gets there too. All $10\times 10\times 10=1000$ digit strings minus the $100$ that start with $0$ is $900$.

Only the first two digits are free, since the third has to copy the second and the fourth has to copy the first. The first digit has $9$ choices because it cannot be $0$, and the second has $10$. Multiply, $9\times 10=90$, so there are $\boxed{{\displaystyle 90}}$ four-digit palindromes. A forced slot has $1$ way, so it contributes a factor of $1$ and never changes the product.

The two ends are forced, the tallest on the left and the shortest on the right, so each end has $1$ way. That leaves $4$ people for the $4$ middle spots, and those fill in $4\times 3\times 2\times 1=24$ ways. The whole row has $\boxed{{\displaystyle 24}}$ orders. Pinning down the forced spots first turns a six-person lineup into a plain arrangement of four.

The first digit must be odd, so it is one of $1,3,5,7,9$, giving $5$ choices. The second must be even, one of $0,2,4,6,8$, another $5$. The third digit has to dodge the $2$ digits already used, leaving $8$, and the fourth dodges $3$, leaving $7$. Multiply, $$5\times 5\times 8\times 7=1400,$$ so $\boxed{{\displaystyle 1400}}$ PINs work. Odd and even digits never overlap, so the second box keeps all $5$ of its choices.

The letter box has $26$ choices. The first digit cannot be $0$, so it has $9$. The second digit can be any digit except the one already used, so $9$ again, and the third must dodge both, leaving $8$. Multiply, $26\times 9=234$, then $234\times 9=2106$, then $2106\times 8=16848$. There are $\boxed{{\displaystyle 16848}}$ tags. Two different restrictions can land on the same box count for different reasons.

One hat pairs with each of the $6$ scarves, so that hat alone gives $6$ pairs. All $3$ hats give three groups of $6$ with no overlap, so $3\times 6=\boxed{{\displaystyle 18}}$ pairs.

There are $2+3=5$ flavors in all. Each of the $5$ cone types pairs with each flavor, so $5\times 5=\boxed{{\displaystyle 25}}$ cones. The flavors arrive in two groups, so add those together first, then multiply across the two stages.

Multiply the options at each step, $2$ drinks, $5$ mains, $3$ fruits. The first two steps give $2\times 5=10$ drink-and-main pairs, and each of those takes any of $3$ fruits, so $10\times 3=\boxed{{\displaystyle 30}}$ combos.

Repeats are allowed, so each of the $3$ lights has all $4$ colors open to it. Multiply, $4\times 4\times 4=4^3=\boxed{{\displaystyle 64}}$ signals. The first light's color does not use anything up, which is why the count stays at $4$ every time.

The leading slot cannot be $0$, so it has the $9$ choices $1$ through $9$. The other three slots each take any digit from $0$ to $9$, so $10$ choices each. Multiply, $$9\times 10\times 10\times 10=9000,$$ so there are $\boxed{{\displaystyle 9000}}$ four-digit numbers.

Fill the five spots left to right. The first has $5$ trophies to choose from, then $4$ are left, then $3$, then $2$, then $1$. Multiply, $5\times 4\times 3\times 2\times 1=120$, so there are $\boxed{{\displaystyle 120}}$ orders. Each trophy placed is used up, and that countdown to $1$ is exactly what $5!$ means.

Think of two spots to fill in order, the gold first and then the silver. For the gold, any of the $8$ runners could finish first, so there are $8$ choices. Once the gold is decided, that runner is taken, which leaves $7$ runners who could come in second for the silver. The multiplication principle says you multiply the number of choices for each spot, and you get $$8\times 7=56.$$ So the two medals can be awarded in $\boxed{{\displaystyle 56}}$ different ways.

The left spot takes any of the $9$ symbols. The right spot can be anything except the symbol you just placed, so $8$ are still allowed. Multiply, $9\times 8=\boxed{{\displaystyle 72}}$ tags. Repeats are allowed in general, and the matching rule bans exactly one symbol on the right, so the count drops by just $1$.

The star painting is locked into the far left spot, so only the other three move. Fill the three open spots one at a time with $3$, then $2$, then $1$ choices, which gives $3\times 2\times 1=\boxed{{\displaystyle 6}}$ orders. A locked spot has $1$ way, so it contributes a factor of $1$ and drops out of the product.

A 4-digit palindrome has the pattern $ABBA$, so only the first two digits are free. No digit may be $0$, so each free digit is one of $1$ through $9$, which is $9$ choices. Multiply, $9\times 9=\boxed{{\displaystyle 81}}$ palindromes. The last two digits are copies, so each contributes a factor of $1$.

It is easier to count the ones we do not want and take them away. There are $900$ three-digit numbers in total, and $90$ of them are palindromes. So the numbers that are not palindromes are everything else, which gives $900-90=810$. So the answer is $\boxed{{\displaystyle 810}}$.

Pick a road from A to B in $3$ ways, then a road from B to C in $4$ ways. Every first road opens all $4$ ways to finish, so multiply, $3\times 4=\boxed{{\displaystyle 12}}$ routes.

Multiply the options at each step, $2$ crusts, $2$ sauces, $3$ cheeses, $4$ toppings. Left to right, $2\times 2=4$, then $4\times 3=12$, then $12\times 4=48$, so there are $\boxed{{\displaystyle 48}}$ pizzas.

The first letter has $26$ choices and the second must differ, leaving $25$, so the letters give $26\times 25=650$. The first digit has $10$ choices and the second $9$, so the digits give $90$. Then $650\times 90=\boxed{{\displaystyle 58500}}$ codes. Making a pair distinct only trims the second slot by one, and everything still multiplies.