Question

Four students (K, L, M, N) participate in four competitions (Quiz, Debate, Chess, Coding), one event each.
1. K does not do Quiz or Chess.
2. L does Coding.
3. N does not do Debate.
4. M does not do the same event type as K.
How many valid assignments are possible?

Hint:

When each person must take a unique role, assign the most constrained individuals first. This reduces the search space dramatically.

Answer 1

Correct Answer: 2

Approach: A one-each bijection between 4 students and 4 events. Start from the most-restricted student (L is fixed; K has only one survivor), then the rest fall out almost forced.

Step 1: L is fixed.
L does Coding. That removes Coding for everyone else. Remaining events: Quiz, Debate, Chess for K, M, N.

Step 2: K is forced.
K \(\ne\) Quiz and K \(\ne\) Chess. Of \(\{\text{Quiz, Debate, Chess}\}\), only Debate survives. So

\[ K = \text{Debate}. \]

Step 3: Clear the trivial rules.
N \(\ne\) Debate is now automatically satisfied (Debate is taken by K). "M does the same event type as K" is impossible anyway in a bijection, so condition 4 adds nothing. Both are non-binding.

Step 4: Split the last two events.
Only Quiz and Chess remain, for M and N. Both orders are legal:

\[ (M=\text{Quiz},\,N=\text{Chess}) \quad\text{or}\quad (M=\text{Chess},\,N=\text{Quiz}). \]

\[ \boxed{2} \]

Final Answer: \(\boxed{2}\)

Intuition: in a small bijection, chase the forced moves first (L, then K). Once two people are locked, the remaining two just swap, giving a clean factor of 2.

Answer 2

Approach: Work by direct elimination on K first, since K has the most restrictions, then see what is left for the rest.

Step 1: K cannot do Quiz or Chess (rule 1). Coding is taken by L (rule 2), so K's only option is Debate. K = Debate.

Step 2: Rule 3 says N does not do Debate; that slot is already taken by K, so this rule adds nothing new for N – it is automatically satisfied.

Step 3: Rule 4 ("M does not do the same event as K") is automatically true for any valid assignment, since each of the four events goes to a different student – M and K can never be given the same event anyway. It carries no extra restriction.

Step 4: Only Quiz and Chess remain, to be split between M and N, in either order.

Final Answer: 2 valid assignments.