A [i]permutation[/i] of a finite set is a one-to-one function from the set onto itself.  A [i]cycle[/i] in a permutation $P$ is a nonempty sequence of distinct items $x_1$, $\ldots\,$, $x_n$ such that $P(x_1) = x_2$, $P(x_2) = x_3$, $\ldots\,$, $P(x_n) = x_1$.  Note that we allow the 1-cycle $x_1$ where $P(x_1) = x_1$ and the 2-cycle $x_1, x_2$ where $P(x_1) = x_2$ and $P(x_2) = x_1$.  Every permutation of a finite set splits the set into a finite number of disjoint cycles.  If this number equals 2, then the permutation is called [i]bi-cyclic[/i].  Compute the number of bi-cyclic permutations of the 7-element set formed by the letters of "PROBLEM".
To find the number of bi-cyclic permutations of the 7-element set formed by the letters of "PROBLEM", we need to count the permutations that split the set into exactly two disjoint cycles. 

1. **Identify the possible cycle structures:**
   - Since we need exactly two disjoint cycles, the possible cycle structures for 7 elements are:
     - One cycle of length 1 and one cycle of length 6.
     - One cycle of length 2 and one cycle of length 5.
     - One cycle of length 3 and one cycle of length 4.

2. **Calculate the number of permutations for each cycle structure:**
   - **Cycle structure (1, 6):**
     - Choose 1 element out of 7 to be in the 1-cycle: $\binom{7}{1} = 7$.
     - The remaining 6 elements form a 6-cycle. The number of ways to arrange 6 elements in a cycle is $(6-1)! = 5! = 120$.
     - Total permutations for this structure: $7 \times 120 = 840$.

   - **Cycle structure (2, 5):**
     - Choose 2 elements out of 7 to be in the 2-cycle: $\binom{7}{2} = 21$.
     - The remaining 5 elements form a 5-cycle. The number of ways to arrange 5 elements in a cycle is $(5-1)! = 4! = 24$.
     - Total permutations for this structure: $21 \times 24 = 504$.

   - **Cycle structure (3, 4):**
     - Choose 3 elements out of 7 to be in the 3-cycle: $\binom{7}{3} = 35$.
     - The remaining 4 elements form a 4-cycle. The number of ways to arrange 4 elements in a cycle is $(4-1)! = 3! = 6$.
     - Total permutations for this structure: $35 \times 6 = 210$.

3. **Sum the permutations for all cycle structures:**
   - Total number of bi-cyclic permutations: $840 + 504 + 210 = 1554$.

The final answer is $\boxed{1554}$