Adding and Subtracting Cases

The floors with people are $3,4,5,6,7,8,9,10,11$, which is $11-3+1=\boxed{{\displaystyle 9}}$ floors. Subtracting alone gives $8$, but that counts the steps you climb between floors. The $3$rd floor never got a step of its own, so add it back.

The announcer says every whole number from $0$ to $30$, both ends included, so the count is $30-0+1=\boxed{{\displaystyle 31}}$. Plain subtraction measures the gaps between numbers, not the numbers themselves. Counting $0$ to $2$ gives $0,1,2$, three numbers, not two.

Between leaves out both ends, so the choices run from $2$ to $49$. That is $49-2+1=\boxed{{\displaystyle 48}}$ whole numbers. The plus one puts back the starting number, which the subtraction skips.

The painted lockers are $5,10,15,\dots ,200$, stepping by $5$. The span is $200-5=195$, which holds $195÷5=39$ steps, so the count is $39+1=\boxed{{\displaystyle 40}}$. The steps are the gaps between lockers, and the first locker takes no step to reach, so it needs its own $+1$.

The ranks played are $96,92,88,\dots ,4$. The span is $96-4=92$ and each step is $4$, so there are $92÷4=23$ steps and $23+1=\boxed{{\displaystyle 24}}$ songs. Counting a falling list works exactly like counting a rising one, and the plus one catches the song you start on.

The sunny-side houses are $24,26,28,\dots ,150$, stepping by $2$. The span is $150-24=126$, so there are $126÷2=63$ gaps and $63+1=\boxed{{\displaystyle 64}}$ houses. After $63$ jumps of $2$ you land on house $150$, and you still have to count house $24$ where you started.

Adding gives $9+15=24$, but you sit in both counts, so drop one copy of yourself. The line holds $9+15-1=\boxed{{\displaystyle 23}}$ people. The front group ends at you and the back group ends at you too, so you get counted twice and the $-1$ undoes it.

If we just add the two groups, $15+12=27$. But that counts the $7$ kids who play both twice, once inside the soccer count and once inside the chess count. To fix the double count we subtract the overlap one time, so $27-7=20$. That gives the number who play at least one game, soccer or chess. $$\boxed{{\displaystyle 20}}$$

Adding the lists gives $22+19=41$, but only $33$ kids signed up at all. The extra $41-33=\boxed{{\displaystyle 8}}$ is what got counted twice. Anyone who picked both archery and swimming sits on each list, so the overcount is exactly the size of the both group.

The coffee-or-tea count is $32+19-8=43$, subtracting the $8$ who landed in both groups. Everyone else drinks neither, so that is $50-43=\boxed{{\displaystyle 7}}$ people. Adding $32+19$ alone gives $51$, more than the whole reunion, which is a signal that the overlap got counted twice.

Sort the plants into four boxes, flowering or not against sunny or shady. The flowering and sunny box holds $22$. Flowering and shady is $40-22=18$, and non-flowering and sunny is $30-22=8$. All four boxes add to $70$, so the non-flowering and shady box holds $70-(22+18+8)=\boxed{{\displaystyle 22}}$. Inclusion-exclusion gives the same number. The plants that are flowering or sunny total $40+30-22=48$, and the rest, $70-48=22$, are in neither group.

The two-digit multiples of $3$ run $12$ to $99$, so there are $(99-12)÷3+1=30$ of them. The two-digit multiples of $5$ run $10$ to $95$, so there are $(95-10)÷5+1=18$. The numbers in both lists are the multiples of $15$, namely $15,30,45,60,75,90$, which is $6$. That gives $30+18-6=\boxed{{\displaystyle 42}}$. Divisible by $3$ and by $5$ means divisible by $15$, and each of those six numbers got counted once in each list, so it comes off once.

Seats $5$ through $23$ number $23-5+1=\boxed{{\displaystyle 19}}$, so $19$ people are sitting. The subtraction alone gives $18$, which counts the steps from the first seat to the last and leaves out one end. Check it on seats $1$ to $3$, where $3-1=2$ but three people are sitting.

One tick per year from $1990$ to $2024$ gives $2024-1990+1=\boxed{{\displaystyle 35}}$ ticks. Subtracting alone counts the gaps between ticks, and a row always has one more tick than gap. From $1990$ to $1992$ there are three ticks, not the two that $1992-1990$ suggests.

The readings run from $-4$ up to $12$. The gap is $12-(-4)=12+4=16$, so the count is $16+1=\boxed{{\displaystyle 17}}$ readings. Subtracting a negative adds, and the plus one is the usual fencepost fix. From $2$ down to $0$ there are three readings even though the gap is $2$.

The restocked slots are $6,12,18,\dots ,96$, since $96=6\times 16$ and the next multiple $102$ is past $100$. The span is $96-6=90$, which holds $90÷6=15$ steps, so the count is $15+1=\boxed{{\displaystyle 16}}$. The steps are the gaps between slots, so add one for slot $6$ itself.

The jerseys step up by $2$, so this is an evenly spaced count. The span is $67-15=52$, which holds $52÷2=26$ steps, and the count is $26+1=\boxed{{\displaystyle 27}}$ jerseys. The steps are the gaps between jerseys, and you are already standing on jersey $15$ before taking any step.

The audited bins are $50,100,150,\dots ,900$. The span is $900-50=850$, which holds $850÷50=17$ steps, so the count is $17+1=\boxed{{\displaystyle 18}}$ bins. Steps count the gaps between bins, and there is always one more bin than gap.

The list drops by $4$ from $100$ down to $20$. The span is $100-20=80$, which holds $80÷4=20$ steps, so the count is $20+1=\boxed{{\displaystyle 21}}$ numbers. Those $20$ steps are gaps between numbers, and $100$ starts the list before any step is taken.

Since $20$ and $95$ are not multiples of $7$, the list is $21,28,35,\dots ,91$. The span is $91-21=70$, which holds $70÷7=10$ jumps, so the count is $10+1=\boxed{{\displaystyle 11}}$. The ten jumps land on the ten numbers after $21$, so $21$ itself needs the extra $+1$.

Adding gives $15+12=27$, but the band sits in both counts. Take off the one duplicate and the parade has $15+12-1=\boxed{{\displaystyle 26}}$ groups. Counting from the start includes the band and counting from the end includes it again, so it lands in the sum twice.

Adding gives $7+7=14$, but you are in both groups, so the line holds $7+7-1=\boxed{{\displaystyle 13}}$ people. Check it straight. Six people ahead of you and six behind you make $6+1+6=13$.

Adding the groups gives $22+18=40$, but the $11$ who ate both sit in each count. Remove one copy and $40-11=\boxed{{\displaystyle 29}}$ students ate at least one. An overlap always gets counted twice, so it comes off once.

Adding the lists gives $28+21=49$, but only $35$ customers bought anything. The excess $49-35=\boxed{{\displaystyle 14}}$ is the number counted twice, which is exactly the both group. Buying both puts a customer on each list, so the sum runs over the true total by the size of the overlap.

Every short-haired entry is either a cat or a dog, so the short-haired cats and short-haired dogs together make $28$. That leaves $28-16=\boxed{{\displaystyle 12}}$ short-haired dogs. The $30$ cats and $20$ dogs are extra facts this question does not need.

The two-digit multiples of $4$ run $12$ to $96$, so there are $(96-12)÷4+1=22$. The two-digit multiples of $9$ run $18$ to $99$, so there are $(99-18)÷9+1=10$. A number divisible by both is divisible by $36$, and the two-digit ones are $36$ and $72$, an overlap of $2$. That gives $22+10-2=\boxed{{\displaystyle 30}}$. Those two numbers appear in each list, so they get counted twice unless you subtract them once.