We have four events: Quiz (Qz), Debate (Db), Chess (Ch), Coding (Cd). Each student must take exactly one unique event.
Step 1: Assign L.
L does Coding. \[ L = Cd \] Remaining events: Qz, Db, Ch.
Remaining students: K, M, N.
Step 2: Apply K’s restriction.
K does not do Quiz or Chess: \[ K \neq Qz,\ K \neq Ch \] Thus the only remaining event K can take is: \[ K = Db \] Step 3: Apply N’s restriction.
N does not do Debate: \[ N \neq Db \] But Db is already taken by K, so N can only take from \[Qz, Ch\].
Step 4: Apply M’s restriction.
M does not do the same event as K.
Since K = Debate, M ≠ Debate.
Remaining options for M are Qz or Ch.
Step 5: Check event uniqueness.
After fixing: \[ L = Cd,\ K = Db \] The two remaining events Qz, Ch must be assigned to M and N in some order. Thus the assignments are: \[ (M = Qz,\ N = Ch) \quad\text{or}\quad (M = Ch,\ N = Qz) \] Both satisfy all constraints.
Total valid assignments: \[ \boxed{2} \] Final Answer: \(\boxed{2}\)
Disregard commonly known facts. Which conclusion would follow on the basis of given statements only?
Statement (I): Some bottles are car. Some cars are cycle.
Conclusion: \[\begin{array}{rl} \bullet & \text{[(I)] Some bottles are cycle is a possibility.} \\ \bullet & \text{[(II)] All bottles are cycle.} \\ \end{array}\]
For any natural number $k$, let $a_k = 3^k$. The smallest natural number $m$ for which \[ (a_1)^1 \times (a_2)^2 \times \dots \times (a_{20})^{20} \;<\; a_{21} \times a_{22} \times \dots \times a_{20+m} \] is: