In a group of 3 girls and 4 boys, there are two boys B_1 and B_2. The number of ways, in which these

Q: Wait, why do we use the complementary method for the boys? Can't we just place them one by one?

You could, but it is a trap! Placing them one by one leads to complex case-by-case analysis that invites errors. Complementary counting is the elegant, 'JEE-proof' way to handle 'not adjacent' constraints. It turns a messy problem into a simple subtraction.

Q: Why do we multiply by $2!$ at the very end?

Because the two blocks—the 'Girls Block' and the 'Boys Block'—are themselves distinct entities. They can swap places: Girls-Boys or Boys-Girls. Ignoring this is a common mistake that costs you half the marks! Always account for the arrangement of the containers themselves.

Q: Does the order of the girls inside their block matter?

Absolutely! The girls are distinct individuals ($G_1, G_2, G_3$). Even though they are a 'block,' they can shuffle their positions within that block in $3!$ ways. Never forget the internal dance of the individuals within the group!

Question

In a group of 3 girls and 4 boys, there are two boys $B_1$ and $B_2$. The number of ways, in which these girls and boys can stand in a queue such that all the girls stand together, all the boys stand together, but $B_1$ and $B_2$ are not adjacent to each other, is:

Options:

(A) 96

(B) 144

(C) 120

(D) 72

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, future engineers. Today, we are not just solving a permutation problem; we are architects of order. We are looking at a group of $3$ girls and $4$ boys, and we need to arrange them in a queue.

We have $3$ girls ($G_1, G_2, G_3$) and $4$ boys ($B_1, B_2, B_3, B_4$). The rules are clear: all girls must stand together, all boys must stand together, and the two boys $B_1$ and $B_2$ must not be adjacent.

### Phase 1: The String Method (Creating the Super-Entities)

Imagine you are the queue manager. To simplify this, we use the 'String Method.' Since all girls must be together, we tie them with an imaginary string and treat them as a single unit, let us call it Unit $G$.

Similarly, we tie all the boys together into Unit $B$. Now, instead of worrying about $7$ individuals, we only have $2$ massive blocks to arrange. This is the beauty of combinatorics—reducing complexity by grouping.

### Phase 2: The Macro Arrangement

With our two units, Unit $G$ and Unit $B$, how many ways can we arrange them? It is a simple binary choice: either the girls stand first, or the boys stand first.

Mathematically, this is $2! = 2$ ways. We have now set the stage for the entire queue, but we must still account for the internal arrangements of these blocks.

### Phase 3: The Internal Dance of the Girls

Inside the girls' block, we have $3$ distinct individuals. They can shuffle their positions among themselves in $3!$ ways.

$$3! = 3 	imes 2 	imes 1 = 6 	ext{ ways}$$

Every time we arrange the girls, we are creating a unique configuration for the queue.

### Phase 4: The Boys' Dilemma (The Core Challenge)

Now, we arrive at the most thrilling part of the problem: the boys. We have $4$ boys, but $B_1$ and $B_2$ refuse to stand next to each other. We use the 'Complementary Counting' strategy to solve this.

First, the total unrestricted arrangements for the $4$ boys is $4! = 24$ ways. Now, let us find the forbidden cases where $B_1$ and $B_2$ are together.

If we force $B_1$ and $B_2$ to be together, we treat them as a single mini-unit. Now we have this mini-unit, plus $B_3$, plus $B_4$, giving us $3$ entities to arrange:

$$3! = 6 	ext{ ways}$$

Within that mini-unit, $B_1$ and $B_2$ can swap places in $2! = 2$ ways. So, the forbidden arrangements are $6 	imes 2 = 12$ ways.

Subtracting the forbidden from the total, we get $24 - 12 = 12$ valid ways for the boys to stand such that $B_1$ and $B_2$ are separated.

### Phase 5: The Final Synthesis

We have all our pieces. The Fundamental Principle of Counting tells us that since these choices are independent, we multiply them. We have the arrangement of the main units ($2!$), the internal arrangement of the girls ($3!$), and the valid internal arrangement of the boys ($12$).

$$Total 	ext{ } Ways = 2! 	imes 3! 	imes 12$$

Substituting the values, we get:

$$2 	imes 6 	imes 12 = 144$$

There you have it! **$144$** distinct ways to form this queue. You have successfully navigated the constraints, used the string method, applied complementary counting, and arrived at the solution.

Select Answer:

Visualized Solution

Understanding the Constraints

The String Method: Creating Units

Arranging the Main Units

Internal Arrangement of Girls

Internal Arrangement of Boys: The Trap

Complementary Counting Strategy

Total Unrestricted Boys Arrangements

Ways $B_{1}$ and $B_{2}$ are Together

Calculating the Forbidden Ways

Valid Boys Arrangements

Final Calculation: The Product Rule

The Final Answer

The Sigma Insight: Linear Permutations

Analyzing the Setup

Phase 1

Phase 2

Phase 3

Phase 4

Phase 5

Similar Questions

The number of ways, 5 boys and 4 girls can sit in a row so that either all the boys sit together or no two boys sit together, is

Two families with three members each and one family with four members are to be seated in a row. In how many ways can they be seated so that the same family members are not separated?

$m$ men and $n$ women are to be seated in a row so that no two women sit together. If $m > n$ , then show that the number of ways in which they can be seated is $\frac{m ! ( m + 1 )!}{( m - n + 1 )!}$

Three persons enter in a lift at the ground floor. The lift will go upto $1 0^{t h}$ floor. The number of ways, in which the three persons can exit the lift at three different floors, if the lift does not stop at first, second and third floors, is equal to ......... .

Eight chairs are numbered 1 to 8. Two women and three men wish to occupy one chair each. First the women choose the chairs from amongst the chairs marked 1 to 4 ; and then the men select the chairs from amongst the remaining. The number of possible arrangements is

The number of ways, in which the letters A, B, C, D, E can be placed in the 8 boxes of the figure below so that no row remains empty and at most one letter can be placed in a box, is :

Total number of ways in which six '+' and four '-' signs can be arranged in a line such that no two '-' signs occur together is .........

How many ways are there to arrange the letters in the word GARDEN with vowels in alphabetical order

The number of arrangements of the letters of the word BANANA in which the two N's do not appear adjacently is