Every Value but One Chosen

Key idea. Conditioning on nobody picking $7$ leaves each person’s choice uniform on the other six values. We then count the maps from seven people onto all six values.

Given that nobody picks $7$, the seven choices are i.i.d. $\mathrm{Unif}\{1,\dots,6\}$. To see this, note that the conditioning event has probability $(6/7)^7$. Each individual assignment avoiding $7$ has probability $7^{-7}$, so its conditional probability is $6^{-7}$. Hence all $6^7$ assignments into $\{1,\dots,6\}$ are equally likely.

We need the number of assignments that use every value in $\{1,\dots,6\}$. Seven people cover six values, so exactly one value is picked by two people and each other value by exactly one. To build such an assignment:

• choose the pair of people who share a value, in $\binom{7}{2}=21$ ways;
• choose their common value, in $6$ ways;
• assign the remaining $5$ values bijectively to the remaining $5$ people, in $5!$ ways.

Each surjection arises exactly once. So there are $21\cdot 6\cdot 5! = 21\cdot 720 = 15120$ of them. Therefore

Alternative. Inclusion–exclusion counts the surjections as $\sum_{j=0}^{6}(-1)^j\binom{6}{j}(6-j)^7$. This also equals $15120$.