Why do we add the results of Case 1 and Case 2?

In combinatorics, 'OR' implies addition for mutually exclusive events. Since the boys cannot be both 'all together' and 'no two together' simultaneously, these events are disjoint, and we sum their individual counts.

Why is it $5! \times 5!$ for the first case?

The first $5!$ accounts for the arrangement of the $5$ units (the 'boy-block' + $4$ girls). The second $5!$ accounts for the internal arrangement of the $5$ boys within that block.

Why are there $5$ gaps in the second case?

When you arrange $4$ girls in a row, they create spaces: one before the first girl, three between the girls, and one after the last girl. That is $4 + 1 = 5$ gaps.

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Animated Solution for Mathematics - Permutations and Combinations: 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

Enter Numerical Value:

Visualized Solution

Understanding the Problem

Total Boys: $5$ , Total Girls: $4$
We have two distinct conditions connected by "OR".
Condition 1: All boys sit together.
Condition 2: No two boys sit together.
These cases are mutually exclusive, so we will add their results.

Case $1$ : All Boys Together

When items must be together, we use the String Method (or Tie Method).
We tie all $5$ boys together and treat them as $1$ single unit.
The $4$ girls remain as $4$ separate units.

Total Units to Arrange

Total units = $1$ (boys unit) + $4$ (girls) = $5$ units.
These $5$ units can be arranged in a row.

Arranging the Units

Number of ways to arrange $5$ units = $5!$
$5! = 5 \times 4 \times 3 \times 2 \times 1 = 120$ ways.

Internal Arrangement of Boys

The $5$ boys inside the "string" can also swap places with each other.
Number of ways to arrange $5$ boys internally = $5! = 120$ ways.

Total Ways for Case $1$

By the Fundamental Principle of Counting, we multiply the independent arrangements.
Total ways for Case 1 = $5! \times 5!$
$120 \times 120 = 14400$ ways.

Case $2$ : No Two Boys Together

When items must be separated, we use the Gap Method.
First, we arrange the other items (the girls) to create gaps.

Arranging the Girls

We have $4$ girls to arrange in a row.
Number of ways to arrange $4$ girls = $4! = 24$ ways.

Identifying the Gaps

The $4$ girls create spaces (gaps) around and between them.
Number of gaps = (Number of girls) + $1$
Total gaps = $4 + 1 = 5$ gaps.

Placing Boys in Gaps

We have $5$ boys and exactly $5$ gaps.
Number of ways to place $5$ boys in $5$ gaps = $^{5} P_{5} = 5!$
$5! = 120$ ways.

Total Ways for Case $2$

Multiply the ways to arrange girls and the ways to place boys.
Total ways for Case 2 = $4! \times 5!$
$24 \times 120 = 2880$ ways.

Final Calculation

The problem asks for Case 1 OR Case 2.
Total ways = (Ways in Case 1) + (Ways in Case 2)
Total ways = $14400 + 2880 = 17280$
Final Answer: $17280$

00:00 / 00:00

The Sigma Insight: Linear Permutations

Solution Diagram

Analyzing the Setup

We are tasked with arranging

5

boys and

4

girls in a row under two specific conditions: either all boys sit together, or no two boys sit together.

The word "OR" indicates that these two scenarios are mutually exclusive. We will calculate the number of arrangements for each case independently and then sum them to find the total.

Case 1

The String Method

When the condition is that all boys must sit together, we treat the

5

boys as a single, inseparable unit.

Now, we have

1

unit of boys and

4

individual girls, totaling

5

units to arrange. These

5

units can be arranged in

5!

ways.

Within the "boy unit," the

5

distinct boys can be arranged among themselves in

5!

ways. By the Fundamental Principle of Counting, the total arrangements for this case are:

14400 = 5! \times 5! = 120 \times 120

Case 2

The Gap Method

When the condition is that no two boys sit together, we utilize the Gap Method. We first arrange the

4

girls, which can be done in

4! = 24

ways.

These

4

girls create

5

available gaps (one at each end and three between the girls). We must place the

5

boys into these

5

gaps.

The number of ways to arrange the

5

boys in these

5

gaps is

5! = 120

. Multiplying these independent choices, we get:

2880 = 4! \times 5! = 24 \times 120

The Final Synthesis

We have explored both mutually exclusive worlds. The first scenario yielded

14400

ways, and the second yielded

2880

ways.

To find the total number of valid arrangements, we add the results from both cases:

14400 + 2880 = 17280

The total number of ways to arrange the group under the given conditions is $17280$ .

Animated Solution for Mathematics - Permutations and Combinations: 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

Enter Numerical Value:

Visualized Solution

Understanding the Problem

Case $1$ : All Boys Together

Total Units to Arrange

Arranging the Units

Internal Arrangement of Boys

Total Ways for Case $1$

Case $2$ : No Two Boys Together

Arranging the Girls

Identifying the Gaps

Placing Boys in Gaps

Total Ways for Case $2$

Final Calculation

The Sigma Insight: Linear Permutations

Analyzing the Setup

Case 1

Case 2

The Final Synthesis

Similar Questions

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:

$m$ 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 $\frac{m ! ( m + 1 )!}{( m - n + 1 )!}$

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?

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

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

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 ......... .

A five-digit numbers divisible by 3 is to be formed using the numerals 0, 1, 2, 3, 4 and 5, without repetition. The total number of ways this can be done is

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Animated Solution for Mathematics - Permutations and Combinations: 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

Enter Numerical Value:

Visualized Solution

Understanding the Problem

Case .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 1: All Boys Together

Total Units to Arrange

Arranging the Units

Internal Arrangement of Boys

Total Ways for Case .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 1

Case .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 2: No Two Boys Together

Arranging the Girls

Identifying the Gaps

Placing Boys in Gaps

Total Ways for Case .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 2

Final Calculation

The Sigma Insight: Linear Permutations

Analyzing the Setup

Case 1

Case 2

The Final Synthesis

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 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 (m−n+1)!m!(m+1)!​

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?

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 .........

A five-digit numbers divisible by 3 is to be formed using the numerals 0, 1, 2, 3, 4 and 5, without repetition. The total number of ways this can be done is

The number of seven digits odd numbers, that can be formed using all the seven digits 1, 2, 2, 2, 3, 3, 5 is ____.

Case $1$ : All Boys Together

Total Ways for Case $1$

Case $2$ : No Two Boys Together

Total Ways for Case $2$

$m$ 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 $\frac{m ! ( m + 1 )!}{( m - n + 1 )!}$