QuanticaPrealgebraOpen in the courseWhen order matters, multiply straight across. But when picking a pair, like two people shaking hands or two teammates, order doesn't matter, and a straight multiply counts every pair twice.
Problem
Six friends each shake hands once with every other friend. How many handshakes happen in all?
Show a hint
- A handshake is a PAIR of friends, and the pair Ana with Ben is the same handshake as Ben with Ana, so the order does not matter. And everyone comes from the one group of six friends, not from two separate groups. So this is a pair count, where you will want to watch out for counting each handshake twice.
- Set up the count the easy way first. Each of the \(6\) friends reaches out to the other \(5\) friends, so multiply to get \(6 \times 5\). Now ask yourself the key question for pairs, did that count each handshake once, or did it sneak each one in twice?
- You have \(6 \times 5 = 30\), but every handshake got counted twice, once from each friend in the pair. So the honest count is half of \(30\), and all that is left is to take that half.
Show the full solution
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 \(\boxed{15}\).
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.
Problem
In an 8-player round-robin tournament, every pair of players meets exactly once. How many matches are played?
Show a hint
- Think about what one match really is. It is a pair of two players, and the match where Ana plays Ben is the same match as the one where Ben plays Ana, so order does not matter. And notice that both players come from the one group of \(8\), not from two separate groups.
- Each of the \(8\) players could face any of the other \(7\), so multiply to get \(8 \times 7\). Now ask the key question, since both players come from the same group, does counting this way count each match once or twice, and if it is twice then you will need to divide by \(2\).
- You have \(8 \times 7 = 56\), and every match got counted twice because each pair shows up from both players' points of view. So take half of \(56\) and you are there.
Show the full solution
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 \(\boxed{28}\) matches.
Both players come from the same group of \(8\), which is exactly when the divide by \(2\) is needed.
Problem
\(4\) students volunteer and exactly \(2\) work together each day. No pair repeats. How many days at most can be scheduled?
Show a hint
- Two students working together is a PAIR, and the order does not matter, since Ana with Ben is the same team as Ben with Ana. And notice all four come from one single group of volunteers, not two separate groups.
- Each student can be paired with the other \(3\), so reach for the multiplying habit and you get \(4 \times 3\). But every pair gets counted twice that way, so think about whether you should divide by \(2\).
- You have \(\frac{4 \times 3}{2}\), which is \(\frac{12}{2}\). The most days is just how many different pairs there are, so finish the division.
Show the full solution
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 \(\boxed{6}\).
Problem
An ice cream shop has \(11\) flavors. A double scoop uses \(2\) different flavors, order doesn't matter. How many different double scoops are possible?
Show a hint
- First decide what kind of count this is. You are choosing \(2\) flavors from one single list of \(11\), and a scoop of chocolate on vanilla is the same scoop as vanilla on chocolate. So this is a pair where the order does not matter, and both flavors come from the same group, not two separate groups.
- Set it up like a handshake. The first flavor has \(11\) choices and the second has \(10\) left, so a first pass gives \(11 \times 10\). But picking chocolate then vanilla and picking vanilla then chocolate land on the same double scoop, so each scoop got counted twice. That tells you whether to divide by \(2\) here.
- You have \(\frac{11 \times 10}{2}\) to finish. Work out the top first, then take half of it, and you have your count.
Show the full solution
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 \(\boxed{55}\) double scoops.
Problem
A dance class has \(7\) leaders and \(6\) followers. Every leader dances once with every follower. How many dances take place?
Show a hint
- Before you count anything, ask what kind of pairing this is. A dance joins one leader with one follower. Are those two coming from the same group, where order would not matter, or are they coming from two different groups you can tell apart?
- You are pulling one pick from the leaders and one pick from the followers, two separate groups, so multiply their sizes. Then pause and decide whether you need to divide by \(2\) the way you do when both picks come from one group.
- Multiplying gives \(7\times6\), and since a leader and a follower can never trade roles, no dance gets counted twice, so there is no dividing by \(2\) here. Just finish that one product.
Show the full solution
A dance takes one leader and one follower, two separate groups, so multiply the sizes and you get $$7\times6=42$$ dances, or \(\boxed{42}\).
There is no dividing by \(2\) here. That fix is only for pairs pulled from one group, where the two picks can trade places.
Problem
At a party, \(5\) women shake hands with every other person, but the \(4\) men never shake hands with each other. How many handshakes take place?
Show a hint
- A handshake is just a pair of people, and the order does not matter, since you shaking my hand is the same single handshake as me shaking yours. So really you are counting how many allowed pairs there are. Notice the pairs come in different kinds, some join two women, some join a woman and a man, and some would join two men.
- Split the count into kinds of pairs. For two women, both come from the same group of \(5\), so that is a pair where you reach out and then divide by \(2\). For a woman and a man, the two picks come from different groups, so you just multiply the sizes and do not divide. For two men, none are allowed, so that kind adds nothing.
- You should have woman-with-woman giving \(\frac{5 \times 4}{2}\), woman-with-man giving \(5 \times 4\), and man-with-man giving \(0\). Work each one out, then add the three counts together.
Show the full solution
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 \(10 + 20 = \boxed{30}\).
When some pairs are off limits, sorting into kinds and counting each with the rule that fits it keeps everything straight.
Problem
A hexagon has \(6\) corners and \(6\) sides. How many diagonals does it have?
Show a hint
- Every line in this picture joins two corners, so you are choosing a pair of corners where the order does not matter. All \(6\) corners come from one single hexagon, not two separate groups, so this is a pairs question and not a multiply-the-two-sizes question.
- First count how many pairs of corners there are at all. Each of the \(6\) corners reaches out to the other \(5\), which gives \(6 \times 5\), and then you divide by \(2\) because joining corner A to corner B is the same line as joining B to A.
- Once you know there are \(15\) pairs of corners in all, remember that some of those lines are sides, not diagonals. The hexagon has \(6\) sides, so take those away from the \(15\) and you are left with just the diagonals.
Show the full solution
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 \(15 - 6 = \boxed{9}\).
Problem
A decagon has \(10\) corners and \(10\) sides. How many diagonals does it have?
Show a hint
- Every diagonal joins two corners, and the order you name them in does not matter, so this is a pair where Ana to Ben is the same as Ben to Ana. And all the corners live in one single group, so you will count pairs from that one group rather than multiply two different sizes.
- Count the pairs of corners first. Each of the \(10\) corners can reach the other \(9\), which gives \(10 \times 9\), and since each pair got counted twice you divide by \(2\). That total includes the sides though, so afterward you will need to take the sides back out.
- You should have \(\frac{10 \times 9}{2} = 45\) pairs of corners. Now just remove the \(10\) that are sides and not diagonals.
Show the full solution
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 \(45 - 10 = \boxed{35}\).
Problem
Six players each play every other player exactly \(3\) games. How many games are played in all?
Show a hint
- First decide what you are really counting. A game is played by a pair of people, and the pair Ana with Ben is the same matchup as Ben with Ana, so order does not matter here. And the two players come from the same group of six, not two separate groups, which is the clue that this is a pair where you will need to undo a double count.
- Count the pairs first. Each of the \(6\) people could face each of the other \(5\), which gives \(6 \times 5\), but that counts every matchup twice since Ana with Ben and Ben with Ana are the same game, so divide by \(2\). Then think about what each pair does after that.
- You have found that there are \(\frac{6 \times 5}{2} = 15\) pairs. Each pair plays \(3\) games, so the last step is to multiply \(15\) by \(3\).
Show the full solution
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 \(15 \times 3 = \boxed{45}\) games.
Count the pairs first, then multiply by how many times each pair does the thing.
Problem
A 17-player round-robin has \(\frac{17 \times 16}{2} = 136\) matches. Verify this by computing the sum \(1+2+3+\cdots+16\) using the Gauss pairing shortcut. What is the value of this sum?
Show a hint
- You are adding up the whole numbers from \(1\) all the way to \(16\). This is the same total as the number of pairs you can choose from \(17\) players, since order does not matter and Ana versus Ben is the same match as Ben versus Ana.
- You could add straight across, but there is a faster way. Try folding the staircase in half. Pair the smallest term with the largest, then the next smallest with the next largest, and notice that every pair makes the same total.
- Pairing the ends gives \(1 + 16 = 17\), \(2 + 15 = 17\), and so on down to \(8 + 9 = 17\). Count how many pairs of \(17\) you have, then multiply.
Show the full solution
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 \(8 \times 17 = \boxed{136}\).
That is the same \(136\) as \(\frac{17 \times 16}{2}\), because counting pairs from \(n\) things is just adding \(1 + 2 + \cdots + (n-1)\).
Problem
A league has \(12\) teams in two divisions of \(6\). Every team plays every division-mate twice and every team in the other division once. How many games are in the full season?
Show a hint
- Sort the games into two kinds, the ones inside a division and the ones across the divisions. Inside a division, a game is a PAIR of teams from one group where order does not matter, so Ana versus Ben is the same game as Ben versus Ana. Across the divisions you are picking one team from each of two groups you can tell apart, which is a different kind of count.
- For one division, first count the pairs with \(\frac{6 \times 5}{2}\), then remember each pair plays twice, and there are two divisions. For the across games, you do not divide by \(2\), since the two teams come from groups you can tell apart, so just multiply the sizes \(6 \times 6\).
- One division gives \(\frac{6 \times 5}{2} = 15\) pairs, doubled is \(30\), and two divisions make \(60\). The across games are \(6 \times 6 = 36\). Now add \(60\) and \(36\).
Show the full solution
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 \(60 + 36 = \boxed{96}\).
Problem
An icosahedron has \(12\) vertices and \(30\) edges. Each face is a triangle. How many interior diagonals does it have?
Show a hint
- A segment just needs two corners, and swapping which corner you name first gives the same segment, so this is a pair where order does not matter, all coming from one group of \(12\) vertices.
- Start by counting every pair of vertices, multiply \(12\) by \(11\) and divide by \(2\) since each segment got named twice. Then think about which of those pairs you do not want, the ones that sit on a common face.
- All pairs come to \(\frac{12 \times 11}{2} = 66\). Every face is a triangle, so the only pairs sharing a face are the \(30\) edges. Now you just have \(66\) and \(30\) waiting to be combined.
Show the full solution
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 \(66 - 30 = \boxed{36}\) 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.
Practice these ideas
Practice
A 13-player round-robin has every pair playing exactly once. How many matches are played?
Show the solution
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 \(\boxed{78}\) matches.
Practice
Fourteen people each shake hands once with every other person. How many handshakes happen?
Show the solution
Each of the \(14\) people shakes hands with the other \(13\), which gives \(14\times13\). Every handshake got counted twice, once from each side, so halve it and you get $$\frac{14\times13}{2}=91,$$ or \(\boxed{91}\) handshakes.
Practice
A team of \(16\) picks \(2\) players as co-captains (they share the same role). How many pairs of co-captains are possible?
Show the solution
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 \(\boxed{120}\).
Practice
Eighteen points lie on a circle. A chord connects each pair of points. How many chords are drawn?
Show the solution
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 \(\boxed{153}\).
Practice
A pentagon has \(5\) corners and \(5\) sides. How many diagonals does it have?
Show the solution
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 \(10 - 5 = \boxed{5}\).
Practice
An octagon has \(8\) corners and \(8\) sides. How many diagonals does it have?
Show the solution
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 \(28 - 8 = \boxed{20}\).
Practice
School A has \(7\) players and School B has \(8\). Every player from A plays every player from B once. How many games are played?
Show the solution
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 \(7 \times 8 = \boxed{56}\) games.
Nothing gets counted twice here, so there is no dividing by \(2\). Halving is only for pairs drawn from a single group.
Practice
At a gathering of \(6\) women and \(3\) men, every woman shakes hands with every other person, but the men never shake hands with each other. How many handshakes happen?
Show the solution
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 \(15 + 18 = \boxed{33}\).
Practice
A club has \(6\) players and every pair plays \(4\) games. How many games are played in all?
Show the solution
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 \(15 \times 4 = \boxed{60}\) games.
Practice
A league has \(8\) teams in two divisions of \(4\). Every team plays each division-mate twice and each team in the other division once. How many games are in the season?
Show the solution
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 \(24 + 16 = \boxed{40}\) games.
Practice
Find the sum \(2 + 4 + 6 + \cdots + 28\) (the first 14 even numbers) using the Gauss pairing trick.
Show the solution
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 \(7 \times 30 = \boxed{210}\).
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.
Practice
A cube has \(8\) vertices, \(12\) edges, and \(6\) square faces (each with \(2\) face diagonals). How many interior diagonals does it have?
Show the solution
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 \(28 - 12 - 12 = \boxed{4}\) interior diagonals.
Practice
A pizza shop offers \(20\) toppings. A 2-topping pizza uses two different toppings, order doesn't matter. How many different 2-topping pizzas are possible?
Show the solution
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 \(\frac{380}{2} = \boxed{190}\).
Practice
A heptagon has \(7\) corners and \(7\) sides. How many diagonals does it have?
Show the solution
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 \(21 - 7 = \boxed{14}\).