Counting Pairs

Each of the $6$ friends shakes hands with the other $5$, which gives $6\times 5=30$. Every handshake got counted twice, once from each side, so take half and you get $\boxed{{\displaystyle 15}}$. The halving is what you owe whenever both people in the pair come from the same group. Ana with Ben and Ben with Ana are one handshake, not two.

Each of the $8$ players can face any of the other $7$, which gives $8\times 7=56$. Every match got counted twice, once from each player's side, so take half and you get $\boxed{{\displaystyle 28}}$ matches. Both players come from the same group of $8$, which is exactly when the divide by $2$ is needed.

No pair repeats, so the most days you can schedule is just the number of different pairs. Each of the $4$ students can partner with the other $3$, which gives $4\times 3=12$, and every pair got counted from both sides, so half of that is $\boxed{{\displaystyle 6}}$.

The first flavor can be any of the $11$ and the second any of the $10$ left, which gives $11\times 10=110$. Chocolate then vanilla is the same dessert as vanilla then chocolate, so every scoop got counted twice and you take half, giving $\boxed{{\displaystyle 55}}$ double scoops.

A dance takes one leader and one follower, two separate groups, so multiply the sizes and you get $$7\times 6=42$$ dances, or $\boxed{{\displaystyle 42}}$. There is no dividing by $2$ here. That fix is only for pairs pulled from one group, where the two picks can trade places.

Count the allowed handshakes by kind. Woman with woman draws both people from the same group of $5$, so that is $\frac{5\times 4}{2}=10$. Woman with man draws from two groups you can tell apart, so no halving, just $5\times 4=20$. Man with man is not allowed, so $0$. Adding the kinds gives $10+20=\boxed{{\displaystyle 30}}$. When some pairs are off limits, sorting into kinds and counting each with the rule that fits it keeps everything straight.

Each of the $6$ corners reaches the other $5$, which gives $6\times 5=30$, and halving the double count leaves $15$ pairs of corners. Every pair is joined by either a side or a diagonal, and $6$ of them are sides, so the diagonals are $15-6=\boxed{{\displaystyle 9}}$.

Each of the $10$ corners reaches the other $9$, and dividing by $2$ for the double count gives $$\frac{10\times 9}{2}=45$$ pairs of corners. Ten of those pairs are the sides, not diagonals, so take them out and you get $45-10=\boxed{{\displaystyle 35}}$.

Each of the $6$ players faces the other $5$, which gives $6\times 5=30$, and halving the double count leaves $15$ matchups. Each matchup plays $3$ games, so the season has $15\times 3=\boxed{{\displaystyle 45}}$ games. Count the pairs first, then multiply by how many times each pair does the thing.

Pair the terms from the ends. $1+16=17$, $2+15=17$, and it keeps going inward to $8+9=17$. The $16$ terms fold into $8$ such pairs, so the sum is $8\times 17=\boxed{{\displaystyle 136}}$. That is the same $136$ as $\frac{17\times 16}{2}$, because counting pairs from $n$ things is just adding $1+2+\cdots +(n-1)$.

Inside one division the number of pairs is $\frac{6\times 5}{2}=15$, and each pair plays twice, so that division has $15\times 2=30$ games. Two divisions give $60$. Across the divisions you take one team from each, two groups you can tell apart, so multiply without halving, $6\times 6=36$. Adding gives $60+36=\boxed{{\displaystyle 96}}$.

All the pairs of vertices come to $$\frac{12\times 11}{2}=66.$$ Every face is a triangle, so the only pairs sharing a face are the $30$ edges. Removing those leaves $66-30=\boxed{{\displaystyle 36}}$ interior diagonals. A triangle has no diagonal of its own, which is why there are no face diagonals to subtract on top of the edges.

Each of the $13$ players faces the other $12$, which gives $13\times 12=156$. Every match got counted twice, once from each player's side, so take half and you get $\boxed{{\displaystyle 78}}$ matches.

Each of the $14$ people shakes hands with the other $13$, which gives $14\times 13$. Every handshake got counted twice, once from each side, so halve it and you get $$\frac{14\times 13}{2}=91,$$ or $\boxed{{\displaystyle 91}}$ handshakes.

The two co-captains share the same role, so naming one first does not make a new choice. Each of the $16$ players could pair with any of the other $15$, which gives $16\times 15=240$, and every pair got counted twice, so half of that is $\boxed{{\displaystyle 120}}$.

A chord is just a pair of points. Each of the $18$ points connects to the other $17$, which gives $18\times 17=306$, and every chord got counted from both ends, so take half and you get $\boxed{{\displaystyle 153}}$.

Each of the $5$ corners pairs with the other $4$, and dividing by $2$ for the double count gives $\frac{5\times 4}{2}=10$ pairs of corners. Five of those pairs are the sides of the pentagon, not diagonals, so the diagonals are $10-5=\boxed{{\displaystyle 5}}$.

Each of the $8$ corners reaches the other $7$, and halving the double count gives $$\frac{8\times 7}{2}=28$$ pairs of corners. Eight of those segments are the sides of the octagon, so take them away and the diagonals are $28-8=\boxed{{\displaystyle 20}}$.

Each game takes one player from School A and one from School B, two groups you can tell apart, so multiply the sizes and you get $7\times 8=\boxed{{\displaystyle 56}}$ games. Nothing gets counted twice here, so there is no dividing by $2$. Halving is only for pairs drawn from a single group.

Woman with woman is a pair from the one group of $6$, so that count is $\frac{6\times 5}{2}=15$. Woman with man takes one from each of two groups you can tell apart, so you just multiply, $6\times 3=18$. The men never shake each other, so nothing more is added. Together that is $15+18=\boxed{{\displaystyle 33}}$.

Each of the $6$ players faces the other $5$, and halving the double count gives $\frac{6\times 5}{2}=15$ pairs. Each pair plays $4$ games, so the club plays $15\times 4=\boxed{{\displaystyle 60}}$ games.

Inside one division the number of pairs is $\frac{4\times 3}{2}=6$, and each pair meets twice, so that division has $6\times 2=12$ games. Two divisions give $24$. Across the divisions you take one team from each, two groups you can tell apart, so multiply without halving, $4\times 4=16$. Adding gives $24+16=\boxed{{\displaystyle 40}}$ games.

Pair the numbers from the ends. $2+28=30$, then $4+26=30$, then $6+24=30$, and the total keeps coming out the same. The $14$ numbers fold into $7$ pairs, so the sum is $7\times 30=\boxed{{\displaystyle 210}}$. Pairing the ends works on any evenly spaced list, since a step in from the left is matched by an equal step in from the right.

All the pairs of corners come to $$\frac{8\times 7}{2}=28.$$ Of those, $12$ are edges and $6\times 2=12$ are face diagonals lying flat on a face. What is left pierces the inside, so $28-12-12=\boxed{{\displaystyle 4}}$ interior diagonals.

Each of the $20$ toppings can join with any of the other $19$, which gives $20\times 19=380$. Mushroom then onion is the same pizza as onion then mushroom, so every pizza got counted twice and you take half, giving $\frac{380}{2}=\boxed{{\displaystyle 190}}$.

Each of the $7$ corners reaches the other $6$, and halving the double count gives $\frac{7\times 6}{2}=21$ pairs of corners. Seven of those lines are the sides of the heptagon, not diagonals, so $21-7=\boxed{{\displaystyle 14}}$.