35. There are $k$ people and $n$ chairs in a row, where $2 \leq k<n$. There is a couple among the $k$ people. The number of ways in which all $k$ people can be seated such that the couple is seated together is equal to the number of ways in which the $(k-2)$ people, without the couple present, can be seated. Find the smallest value of $n$.
35. Answer: 12 .
$$
(n-1) \times 2 \times\binom{ n-2}{k-2} \times(k-2)!=\binom{n}{k-2} \times(k-2)!\text {. Then } 2=\frac{n}{(n-k+1)(n-k+2)} .
$$

That is, $2 n^{2}-(4 k-6) n+\left(2 k^{2}-6 k+4-n\right)=0$ We can solve
$$
n=k+\frac{\sqrt{8 k-7}-5}{4} \quad(n>k) .
$$

Note that the square of any odd number has the form $8 k-7$. Choose $k$ so that $\sqrt{8 k-7}-5=4$, i.e., $k=11$. Then $n=12$.