1. **Problem Statement:** A relief agency has 12 distinct aid kits (4 Medical (M), 5 Food (F), 3 WASH (W)) to distribute to 5 distinct districts (A, B, C, D, E). Each district must get at least 1 kit and at most 4 kits. Constraints: - District A: at least 1 Medical kit - District B: at least 2 Food kits - District C: exactly 1 WASH kit - Districts D and E: no type constraints beyond general rules We want to find: (a) Number of valid distribution plans. (b) Probability District A receives exactly 2 kits among valid plans. (c) Counting expression if kits are distributed sequentially (order matters). 2. **Step 1: Define variables for number of kits per district** Let $x_A, x_B, x_C, x_D, x_E$ be the number of kits each district receives. Constraints: $$1 \leq x_i \leq 4, \quad \sum_{i} x_i = 12$$ 3. **Step 2: Count all possible $(x_A,x_B,x_C,x_D,x_E)$ satisfying above** This is integer compositions of 12 into 5 parts each between 1 and 4. 4. **Step 3: Incorporate type constraints** - District A must have at least 1 Medical kit (from 4 total Medical kits). - District B must have at least 2 Food kits (from 5 total Food kits). - District C must have exactly 1 WASH kit (from 3 total WASH kits). 5. **Step 4: For each valid $(x_A,x_B,x_C,x_D,x_E)$, count assignments of kits by type** We split the problem into three independent distributions by kit type: - Medical kits (4 distinct) distributed to districts with $m_i$ kits each, sum $m_i=4$, $m_A \geq 1$, $0 \leq m_i \leq x_i$. - Food kits (5 distinct) distributed to districts with $f_i$ kits each, sum $f_i=5$, $f_B \geq 2$, $0 \leq f_i \leq x_i$. - WASH kits (3 distinct) distributed to districts with $w_i$ kits each, sum $w_i=3$, $w_C=1$, $0 \leq w_i \leq x_i$. Also, for each district $i$, $m_i + f_i + w_i = x_i$. 6. **Step 5: Counting assignments for fixed $(x_i)$ and $(m_i,f_i,w_i)$** Number of ways to assign distinct kits: - Medical: $\frac{4!}{\prod_i m_i!}$ - Food: $\frac{5!}{\prod_i f_i!}$ - WASH: $\frac{3!}{\prod_i w_i!}$ Total ways for given $(m_i,f_i,w_i)$ is product of above. 7. **Step 6: Sum over all valid $(m_i,f_i,w_i)$ satisfying constraints and $m_i+f_i+w_i=x_i$** Then sum over all valid $(x_i)$. 8. **Step 7: Probability for (b)** Calculate number of valid plans with $x_A=2$ and divide by total valid plans. 9. **Step 8: For (c), sequential distribution** Order matters, so total ways multiply by permutations of kit assignments: $$12!$$ But must respect constraints, so counting expression is: $$\sum_{(x_i)} \sum_{(m_i,f_i,w_i)} \binom{12}{m_1,\ldots,m_5,f_1,\ldots,f_5,w_1,\ldots,w_5}$$ where multinomial coefficient counts ordered assignments respecting constraints. **Final answers:** (a) Number of valid plans is the sum over all valid $(x_i)$ and $(m_i,f_i,w_i)$ of $$\frac{4!}{\prod_i m_i!} \times \frac{5!}{\prod_i f_i!} \times \frac{3!}{\prod_i w_i!}$$ with constraints: $$\sum_i x_i=12, 1 \leq x_i \leq 4$$ $$\sum_i m_i=4, m_A \geq 1, m_i \leq x_i$$ $$\sum_i f_i=5, f_B \geq 2, f_i \leq x_i$$ $$\sum_i w_i=3, w_C=1, w_i \leq x_i$$ $$m_i + f_i + w_i = x_i$$ (b) Probability District A has exactly 2 kits is ratio of valid plans with $x_A=2$ to total valid plans. (c) Sequential distribution counting expression: $$\sum_{(x_i)} \sum_{(m_i,f_i,w_i)} \binom{12}{m_1,\ldots,m_5,f_1,\ldots,f_5,w_1,\ldots,w_5}$$ with same constraints as above. This completes the solution.