Problem 1695

1. Define the problem and variables:
Let $n$ be the total number of people in the group. Each person has exactly $d$ friends, and every two strangers have exactly one common friend.

2. **Define the set $S$:**
Let $S$ be the set of ordered triples of distinct people $(A, B, C)$ such that $A$ knows $B$, $B$ knows $C$, but $C$ does not know $A$.

3. **Count the size of $S$ in two ways:

- First way (based on $A$):**
For each person $A$, there are $n - d - 1$ possibilities for $C$ (since $A$ has $d$ friends and there are $n - 1$ other people, so $n - d - 1$ are strangers to $A$). Each such $C$ has exactly one common friend $B$ with $A$ (by the problem's condition). Therefore, the number of such triples for each $A$ is $n - d - 1$. Since there are $n$ people, we have:
$$|S| = n(n - d - 1)$$

- **Second way (based on $B$):**
For each person $B$, we choose $A$ and $C$ out of the $d$ friends of $B$. There are $d$ choices for $A$ and $d - 1$ remaining choices for $C$ (since $A \neq C$). Therefore, the number of such triples for each $B$ is $d(d - 1)$. Since there are $n$ people, we have:
$$|S| \leq nd(d - 1)$$

4. Combine the two counts:
Equating the two expressions for $|S|$, we get:
$$n(n - d - 1) \leq nd(d - 1)$$

5. Simplify the inequality:
$$n(n - d - 1) \leq nd(d - 1)$$
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