Animated Solution for Mathematics - Permutations and Combinations: There are ten boys $B_1,B_2,\dots ,B_{10}$ and five girls $G_1,G_2,\dots ,G_5$ in a class. Then the number of ways of forming a group consisting of three boys and three girls, if both $B_1$ and $B_2$ together should not be the members of a group, is ______.

Mastering the Constraint

Welcome, future engineer. Today, we are not just solving a combinatorics problem; we are learning the philosophy of selection. In the JEE Advanced arena, you will often face problems that seem straightforward but contain hidden traps.

The problem of selecting students from a class with a specific constraint is a classic test of your logical architecture. Let us break this down, step by step, and turn this complexity into clarity.

The Setup

Imagine you are standing in front of a class of fifteen students: ten boys ($B_1,B_2,\dots ,B_{10}$) and five girls ($G_1,G_2,\dots ,G_5$). Your mission is to form a committee of six: exactly three boys and three girls.

In the world of combinatorics, we use the combination formula:

Our goal is to select $3$ boys from $10$ and $3$ girls from $5$. If there were no rules, the math would be simple: ${}^{10}{C}_3{\times }^5{C}_3$. But life—and JEE problems—rarely gives us such freedom.

The Philosophy of Complementary Counting

We have a constraint: $B_1$ and $B_2$ cannot be in the same group. They are the 'forbidden pair.'

When you see a 'not together' constraint, your first instinct should be the Complementary Method. Think of it this way: the total number of ways to form a group is the sum of all possible combinations.

This total universe consists of two types of groups: those where $B_1$ and $B_2$ are together (the forbidden ones) and those where they are not (the valid ones).

Instead of trying to count the valid ones directly, we take the shortcut:

This is the mark of an efficient problem solver. Why work harder when you can work smarter?

Calculating the Total Universe

First, let us calculate the total unrestricted ways. We need to select $3$ boys from $10$ and $3$ girls from $5$.

For the boys:

For the girls:

Multiplying these together gives us the total number of ways to form the group without any constraints:

This is our baseline. This is the number of ways we could form the group if $B_1$ and $B_2$ were the best of friends.

The Forbidden Zone

Now, we must identify the 'Forbidden Ways.' These are the groups where $B_1$ and $B_2$ are both present.

Imagine you have already placed $B_1$ and $B_2$ into the group. You have effectively filled $2$ of the $3$ boy-slots. You only have $1$ slot left for the boys.

Since $2$ are already taken, we have $10-2=8$ boys remaining. The number of ways to fill the remaining boy-slot is:

What about the girls? The constraint does not mention them, so they remain free. We still need to select $3$ girls from $5$:

Therefore, the number of forbidden groups is:

The Final Synthesis

We have arrived at the final moment. We have our total universe ($1200$) and our forbidden scenarios ($80$). To find the number of valid groups where $B_1$ and $B_2$ are not together, we simply subtract:

There it is. The final answer is 1120.

Do you see the elegance? We didn't need to struggle with complex cases. By identifying the constraint, applying the complementary method, and carefully calculating the forbidden subset, we navigated the problem with precision.

This is the mindset you need for the JEE Advanced. It is not about memorizing formulas; it is about visualizing the logic, respecting the constraints, and executing the math with confidence. Keep practicing this logical flow, and you will find that even the most daunting problems become solvable puzzles.