In a small village, there are 87 families, of which 52 families have at most 2 children. In a rural development program 20 families are to be chosen for assistance, of which at least 18 families must have at most 2 children. In how many ways can the choice be made? (A) 52C18 × 52C2 + 52C19 × 52C1 (B) 52C18 × 35C2 + 52C19 × 35C1 + 52C20 (C) 52C18 × 52C2 + 52C19 × 52C1 + 52C20 (D) 52C18 × 35C2 + 52C19 × 35C1

Key Concepts Needed

1. Binomial (Combination) Formula
   The number of ways to choose r objects from n distinct objects without caring about order is

2. At Least / At Most Conditions
   "At least 18 families with at most 2 children" means acceptable selections must satisfy 18, 19, or 20 such families.

3. Partitioning the Sample Space
   Break the big counting job into exclusive cases (they don’t overlap) and then add the counts.

4. Complement Family Counts
   Total families = 87. Given 52 families have "≤2 kids", automatically 87−52 = 35 families have ">2 kids".
   This second group will fill the remaining slots when fewer than 20 "≤2-kids" families are chosen.

Logical Chain to Solve

Step-1: Identify the two pools: 52 ‘eligible’ (≤2 kids) and 35 ‘others’.
Step-2: List all selection patterns that satisfy “at least 18 eligible”.
Step-3: For each pattern apply $\binom{n}{r}$ separately to the two pools and multiply (because you do one choice and the other).
Step-4: Add the three independent case counts to get the final answer.

This systematic approach guarantees you haven’t double-counted or missed anything.

Imagine Picking Families Like Picking Mangoes

Think of 87 mangoes in two baskets:

• Basket-A has 52 sweet mangoes (families with at most 2 kids).
• Basket-B has 35 sour mangoes (families with more than 2 kids).

You have to pack 20 mangoes for a gift box, but at least 18 of them must be sweet.
So your box can look only like:

1. 18 sweet + 2 sour
2. 19 sweet + 1 sour
3. 20 sweet + 0 sour

For each pattern you first decide how many from Basket-A and then how many from Basket-B.
Mathematically we use a special counting button called “n choose r” (written $\binom{n}{r}$) which tells us how many ways to pick r items from n items.

Add the counts for the three patterns and you get the total number of ways to make the gift box.

Step-by-Step Solution

Let

• $N = 87$ total families
• $E = 52$ families with at most 2 children ("eligible")
• $O = N-E = 35$ families with more than 2 children ("others")
• Required: Select 20 families such that $\ge 18$ of them are from $E$

We enumerate the valid cases.

Case-1: 18 Eligible, 2 Others

Eligible choices:

Other choices:

Total for this case:

Case-2: 19 Eligible, 1 Other

Case-3: 20 Eligible, 0 Others

Add the Three Cases

Comparing with the given options, this matches Option (B).

[ \boxed{\text{Option (B)}} ]