Problem statement
Five schools are sending their baseball teams to a tournament. Each team must play every other team exactly once. How many games are required in total?
General form. Given a set $S = \{s_1, s_2, \dots, s_n\}$ of $n$ distinct elements, how many unordered pairs $\{s_i, s_j\}$ with $s_i \neq s_j$ can be formed?
We solve this three ways. Each approach reveals a different combinatorial principle, and all three land on the same answer: \[ \boxed{\;\dfrac{n(n-1)}{2}\;=\;\binom{n}{2}\;} \]
Approach 1: The concrete case $n=5$
Lay out a $5 \times 5$ table whose rows and columns are indexed by the teams $T_1, \dots, T_5$. A check mark in row $T_i$, column $T_j$ records the game between team $i$ and team $j$.
Reading the table.
- Diagonal ($\times$, red): a team cannot play itself.
- Lower triangle ($\times$, gray): the pair $(T_i, T_j)$ with $i>j$ records the same game as $(T_j, T_i)$, so we strike it out to avoid double-counting.
- Upper triangle ($\checkmark$, green): exactly one entry per distinct unordered pair.
Counting the checks row by row gives \[ 4 + 3 + 2 + 1 + 0 \;=\; 10 \text{ games.} \]
Approach 2: Row-by-row generalization
The same table works for any $n$. In row $T_k$, the cell in column $T_j$ is a check exactly when $j > k$, so row $T_k$ contributes $n-k$ unordered pairs.
For row $T_k$, any element up to and including $T_k$ itself cannot be considered: the diagonal cell ($T_k$ playing itself) is excluded, and all cells with $j < k$ are already accounted for in earlier rows. Only the $n-k$ entries with $j > k$ contribute.
Summing over all rows: \[ \begin{aligned} \text{Total} \;=\; \sum_{k=1}^{n}(n-k) &\;=\; \underbrace{(n-1) + (n-2) + \cdots + 1 + 0}_{n \text{ terms}} \\ &\;=\; \sum_{k=1}^{n} n \;-\; \sum_{k=1}^{n} k \\ &\;=\; n\cdot n \;-\; \frac{n(n+1)}{2} \\ &\;=\; \frac{2n^2 - n(n+1)}{2} \;=\; \frac{n(n-1)}{2}. \end{aligned} \]
Sanity check. For $n=5$: $\dfrac{5\cdot 4}{2} = 10$. ✓
Approach 3: Count ordered pairs, then divide
Forget the table. Think of building a pair by filling two labeled slots in sequence: \[ \bigl(\;\underline{\;\text{1st}\;}\;,\;\underline{\;\text{2nd}\;}\;\bigr). \]
- The first slot can be filled in $n$ ways (any element of $S$).
- Once the first slot is fixed, the second slot can be filled in $n-1$ ways: any element except the one already chosen (since the pair must contain two different elements).
Therefore the number of ordered pairs of distinct elements is \[ n(n-1). \]
Each unordered pair $\{s_i, s_j\}$ corresponds to exactly two ordered pairs, namely $(s_i, s_j)$ and $(s_j, s_i)$. The count $n(n-1)$ therefore double-counts every unordered pair, and we recover the unordered count by dividing by the size of the symmetry group ($2$): \[ \text{unordered pairs} \;=\; \frac{n(n-1)}{2}. \]
This is a special case of a general rule: when every object is counted the same number of times $m$, divide by $m$.
Result
\[ \binom{n}{2} \;=\; \frac{n(n-1)}{2}. \]
For the baseball tournament with $n=5$ teams: \[ \binom{5}{2} \;=\; \frac{5\cdot 4}{2} \;=\; 10 \text{ games.} \]
Cross-check: small values
| $n$ | Row-sum $\sum_{k=1}^{n}(n-k)$ | Ordered$/2 = n(n-1)/2$ | $\binom{n}{2}$ |
|---|---|---|---|
| $2$ | $1+0 = 1$ | $2\cdot 1/2 = 1$ | $1$ |
| $3$ | $2+1+0 = 3$ | $3\cdot 2/2 = 3$ | $3$ |
| $4$ | $3+2+1+0 = 6$ | $4\cdot 3/2 = 6$ | $6$ |
| $5$ | $4+3+2+1+0 = 10$ | $5\cdot 4/2 = 10$ | $10$ |
| $6$ | $5+4+3+2+1+0 = 15$ | $6\cdot 5/2 = 15$ | $15$ |
All three columns agree, as they must.
Conclusion
The number of unordered pairs of distinct elements drawn from an $n$-element set is \[ \binom{n}{2} \;=\; \frac{n(n-1)}{2}. \] We proved this three ways:
- Concrete table ($n=5$): count check marks in the upper triangle.
- Row-by-row sum: row $T_k$ contributes $n-k$ pairs; summing yields $n^2 - \tfrac{n(n+1)}{2} = \tfrac{n(n-1)}{2}$.
- Division principle: $n(n-1)$ ordered pairs, each unordered pair counted twice, so divide by $2$.
The three viewpoints—enumerate carefully, sum a pattern, overcount and divide—are the standard combinatorial moves and recur throughout the subject.
Remark (where this quantity shows up). The number $\binom{n}{2}$ counts:
- the edges of the complete graph $K_n$ on $n$ vertices;
- the handshakes exchanged when $n$ people each shake hands with every other person once;
- the number of $2$-element subsets of an $n$-set;
- the $n$-th triangular number $T_{n-1} = 1 + 2 + \cdots + (n-1)$.
Remark (generalization). The third approach extends immediately: the number of unordered $r$-subsets of an $n$-set is \[ \binom{n}{r} \;=\; \frac{n(n-1)(n-2)\cdots(n-r+1)}{r!}, \] obtained by counting ordered $r$-tuples of distinct elements ($n(n-1)\cdots(n-r+1)$ of them) and dividing by the number of orderings of each $r$-subset ($r!$). For $r=2$ this recovers $\tfrac{n(n-1)}{2}$.
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