Casework

There are $5$ tacos, one for each filling. A burrito needs a filling and a salsa, so there are $4\times 2=8$ burritos. Nothing is both, so add the two piles, $$5+8=13.$$ You can order $\boxed{{\displaystyle 13}}$ different items. One multiply cannot handle this in one shot, since a taco makes one choice and a burrito makes two. Multiply inside a case, add across cases.

Count by length. One-letter words give $6$, two-letter words give $6^2=36$, and three-letter words give $6^3=216$. A word has exactly one length, so the piles do not overlap and you add, $$6+36+216=258.$$ There are $\boxed{{\displaystyle 258}}$ possible words. Inside a case every slot has $6$ choices, so you multiply. Across the cases you add.

From $D$ to $H$ there are $3$ routes, one through $E$, one through $F$, and the direct road. Each of the $4$ routes from $A$ to $D$ can be followed by any of those, so $$4\times 3=12.$$ There are $\boxed{{\displaystyle 12}}$ routes from $A$ to $H$. Every route passes through $D$, which is why the trip splits there. The two halves are independent, so this step is a multiply, not an add.

Split by how many digits a number has. The numbers $1$ to $9$ are $9$ numbers with $1$ digit each, so $9\times 1=9$. The numbers $10$ to $99$ are $90$ numbers with $2$ digits each, so $90\times 2=180$. The numbers $100$ to $150$ are $51$ numbers with $3$ digits each, so $51\times 3=153$. Adding the piles, $$9+180+153=342,$$ so you write $\boxed{{\displaystyle 342}}$ digits. The piles never overlap and together cover every number, so nothing is counted twice and nothing slips through.

Case on the tens digit $t$. The units digit only has to be smaller, so it can be $0,1,\dots ,t-1$, which is $t$ choices. Running $t$ from $1$ to $9$ and adding gives $$1+2+3+4+5+6+7+8+9=45,$$ so there are $\boxed{{\displaystyle 45}}$ such numbers. You cannot just multiply $9\times 10$ here, because how many units digits are allowed depends on which tens digit you picked.

Split by hat. Hat 1 holds $15$ balls, so the two draws give $15\times 14=210$. Hat 2 holds $10$ balls, so $10\times 9=90$. A draw comes from exactly one hat, so add, $$210+90=300.$$ That is $\boxed{{\displaystyle 300}}$ ordered draws. The second factor drops by one because the first ball stays out. The hats hold different amounts, which is why one multiply cannot cover both at once.

Case on the number of dimes. With $0$ dimes there is $20$ cents to fill, so the nickels can be $0,1,2,3,$ or $4$ and pennies fill the rest, which is $5$ ways. With $1$ dime there is $10$ cents left, so nickels can be $0,1,$ or $2$, which is $3$ ways. With $2$ dimes nothing is left, which is $1$ way. Three dimes is already $30$ cents, too much. Adding, $$5+3+1=9,$$ so there are $\boxed{{\displaystyle 9}}$ ways. Split on the coin that boxes you in the most, here the dime. Once the dimes and nickels are set, the pennies are forced, so there is nothing left to choose.

Case on $a$. With $a=1$ we need $1+b<24$, so $b<23$ and $b$ runs $1$ to $22$, which is $22$ pairs. With $a=2$, $b<20$ gives $19$ pairs. With $a=3$, $b<15$ gives $14$. With $a=4$, $b<8$ gives $7$. At $a=5$ we already have $a^2=25$, too big on its own, so we stop. Adding, $$22+19+14+7=62,$$ so there are $\boxed{{\displaystyle 62}}$ pairs. The room left for $b$ shrinks as $a$ grows, so no single multiply works. Watch the endpoint too, since $b<23$ means $22$ values, not $23$.

Two digits add to somewhere between $1$ and $18$, so the perfect squares in reach are $1$, $4$, $9$, and $16$. Sum $1$ gives only $10$, so $1$ number. Sum $4$ gives $13$, $22$, $31$, $40$, so $4$. Sum $9$ gives $18$, $27$, $36$, $45$, $54$, $63$, $72$, $81$, $90$, so $9$. Sum $16$ gives $79$, $88$, $97$, so $3$. Adding, $$1+4+9+3=17,$$ so there are $\boxed{{\displaystyle 17}}$ numbers. Pinning down which sums are even reachable is most of the work. Miss one square and the count comes up short.

The hundreds digit is never $0$, since $061$ is really just $61$. So the lone $0$ sits in the tens place or the units place. With the $0$ in the tens place, the hundreds digit has $9$ choices and the units digit has $9$, giving $9\times 9=81$. The $0$ in the units place works the same way, another $81$. Adding, $$81+81=162,$$ so there are $\boxed{{\displaystyle 162}}$ numbers. The other free digit has to be nonzero, or you would have two zeros instead of exactly one.

Case on the tens digit $t$. The units digit never appears in the rule, so it always brings $10$ choices. For $t=0$ the hundreds digit needs to be at least $0$, so all $9$ work. For $t=1$ it must be at least $3$, so $7$ work. For $t=2$, at least $6$, so $4$. For $t=3$, at least $9$, so $1$. For $t=4$ or more you would need a digit of $12$ or bigger, so those piles are empty. That gives $$9+7+4+1=21$$ hundreds-tens pairs, each carrying $10$ units digits, so $$21\times 10=210.$$ The count is $\boxed{{\displaystyle 210}}$. A digit no rule touches just multiplies in at the end. Pull it out early and the casework gets much shorter.

Count the numbers with no $2$ instead. Write every number as three slots, so $7$ is $007$ and $52$ is $052$. A slot dodging the $2$ has $9$ choices, giving $9\times 9\times 9=729$ strings from $000$ to $999$. Drop $000$, which stands for $0$ and is not in the range, leaving $728$. There are $999$ numbers in all, so $$999-728=271,$$ and $\boxed{{\displaystyle 271}}$ of them have at least one $2$. Counting the $2$s head on means overlapping cases, since a number can carry two or three of them. Counting the other side avoids all of that.

A small sundae is just a flavor, so there are $5$ of them. A large is a flavor and a sauce, so $5\times 2=10$. No sundae is both sizes, so add, $$5+10=15.$$ There are $\boxed{{\displaystyle 15}}$ different sundaes. Multiply inside a case, add across the cases.

Split by length. A $1$-symbol code has one slot with $5$ shapes, so $5$ codes. A $2$-symbol code has two slots with $5$ shapes each, so $5^2=25$ codes. A code has exactly one length, so add, $$5+25=30.$$ There are $\boxed{{\displaystyle 30}}$ possible codes. When the things you are counting come in different lengths, one multiply cannot reach them all. Split by length first.

Split by how many digits a number has. The numbers $1$ to $9$ are $9$ numbers with one digit each, so $9$ digits. The numbers $10$ to $99$ are $90$ numbers with two digits each, so $90\times 2=180$ digits. Adding, $$9+180=189,$$ so $\boxed{{\displaystyle 189}}$ digits are written. A $7$ costs one digit and a $53$ costs two, so there is no single length to multiply by.

Case on the number of dimes, which can be $0$, $1$, $2$, or $3$. Once the dimes and nickels are chosen the pennies are forced, so only the nickels need counting. With $30$ cents left you can use $0$ up to $6$ nickels, which is $7$ ways. With $20$ cents left, $0$ up to $4$ nickels, which is $5$ ways. With $10$ cents left, $3$ ways. With $0$ cents left, $1$ way. Adding, $$7+5+3+1=16,$$ so there are $\boxed{{\displaystyle 16}}$ ways. Split on the biggest coin. It has the fewest possible values, so it pins the rest down fastest.

Case on $a$. Since $b$ is at least $1$, we need $a^2<14$, so $a$ can only be $1$, $2$, or $3$. With $a=1$ we need $b<14$, so $b$ runs $1$ to $13$, which is $13$ pairs. With $a=2$, $b<11$ gives $10$ pairs. With $a=3$, $b<6$ gives $5$ pairs. Adding, $$13+10+5=28,$$ so there are $\boxed{{\displaystyle 28}}$ pairs. Checking which $a$ are even possible first keeps the case list short and stops you from hunting past $a=3$.

Case on the tens digit $t$. The units digit must be bigger, so it runs from $t+1$ up to $9$, which is $9-t$ choices. That gives $8$ when $t=1$, $7$ when $t=2$, and so on down to $1$ when $t=8$, with nothing left at $t=9$. Adding, $$8+7+6+5+4+3+2+1=36,$$ so there are $\boxed{{\displaystyle 36}}$ such numbers. Each case count shrinks by one, so the whole total is just $1$ through $8$ added up.

Case on the tens digit $t$, with the units digit free at $10$ choices throughout. We need the hundreds digit $h$ from $1$ to $9$ with $h\ge 2t$. For $t=0$ all $9$ work. For $t=1$ we need $h\ge 2$, so $8$ work. For $t=2$ we need $h\ge 4$, so $6$. For $t=3$ we need $h\ge 6$, so $4$. For $t=4$ we need $h\ge 8$, so $2$. From $t=5$ on you would need $h\ge 10$, which no digit reaches. So $$(9+8+6+4+2)\times 10=29\times 10=290,$$ giving $\boxed{{\displaystyle 290}}$ numbers. The units digit never appears in the rule, so pull its $10$ choices out and multiply them in at the end.

Case on the hundreds digit $h$, which runs from $1$ to $9$. With $h$ fixed, the tens digit can be anything from $0$ up to $h$ and the units digit is forced to be whatever is left, so that is $h+1$ ways. Adding over $h=1$ through $h=9$, $$2+3+4+\cdots +10=54,$$ so there are $\boxed{{\displaystyle 54}}$ numbers. Split on the digit the rule pins down. Here $h$ sets the target sum, and then one choice forces the other.

Split by box. Box 1 has $8$ cards, so $8\times 7=56$ ordered draws. Box 2 has $6$ cards, so $6\times 5=30$. A draw comes from one box only, so add, $$56+30=86.$$ There are $\boxed{{\displaystyle 86}}$ ordered draws. The second factor drops by one because the first card is not put back.

The thousands digit is never $0$, so the lone $0$ sits in the hundreds, tens, or ones spot, which gives $3$ cases. Take the $0$ in the hundreds spot. The thousands digit has $9$ choices, and the tens and ones must each be nonzero for $9$ choices each, so that case is $9\times 9\times 9=729$. The other two cases work out the same, so $$3\times 729=2187,$$ giving $\boxed{{\displaystyle 2187}}$ numbers. A number lands in one case only, decided by where its single $0$ sits. When every case is the same size, adding them is just multiplying by the number of cases.

Count the numbers with no $7$ instead. Write each number from $0$ to $499$ as three slots. The hundreds slot is $0$ through $4$ and none of those is a $7$, so $5$ choices, while the tens and ones slots each have $9$ choices once $7$ is banned. That is $5\times 9\times 9=405$ strings. One of them is $000$, which is outside our range, but $500$ has no $7$ and belongs in the pile, so the two adjustments cancel and $$405-1+1=405$$ numbers from $1$ to $500$ avoid the digit $7$. Everything else has one, so $$500-405=95,$$ giving $\boxed{{\displaystyle 95}}$ numbers. Counting the $7$s head on means overlapping cases, since $77$ would get caught twice. Counting the other side sidesteps that.

Split on which spot holds the digit, which gives $3$ cases. With the digit first, that spot has $10$ choices and the two letter spots have $26$ each, so $10\times 26\times 26=6760$ codes. The digit in the second or third spot gives $6760$ each as well, so $$3\times 6760=20280,$$ giving $\boxed{{\displaystyle 20280}}$ codes. Every code has its digit in exactly one spot, so the three cases do not overlap and nothing is counted twice.

Case on the hundreds digit $h$, which must be at least $1$ and at most $4$ or the sum overshoots. With $h$ fixed, the tens and units digits must add to $4-h$, and a target sum of $s$ allows $s+1$ pairs, since the tens digit runs $0$ up to $s$ and the units digit is forced. So $h=1$ gives $4$ ways, $h=2$ gives $3$, $h=3$ gives $2$, and $h=4$ gives $1$. Adding, $$4+3+2+1=10,$$ so there are $\boxed{{\displaystyle 10}}$ numbers. Split on the most cramped digit. The hundreds digit is squeezed from both sides here, which keeps the case list to four.

Case on $b$, since the rule rearranges to $a<20-b^2$. With $b=1$ we need $a<19$, so $18$ values. With $b=2$, $a<16$ gives $15$. With $b=3$, $a<11$ gives $10$. With $b=4$, $a<4$ gives $3$. At $b=5$ we have $b^2=25$, already too big on its own, so the cases stop. Adding, $$18+15+10+3=46,$$ so there are $\boxed{{\displaystyle 46}}$ pairs. Split on the squared variable, because it is the one that eats the room fastest and so gives the fewest cases.