Two families with three members each and one family with four members are to be seated in a row. In

Q: Wait, why do we multiply the internal arrangements? Shouldn't we add them because they are different families?

That is a fantastic question! We multiply because these events are independent stages of a single, larger process. Think of it this way: for every single way you arrange the families (the blocks), you have a set number of ways to arrange the members inside. If you have $X$ ways to arrange the blocks and $Y$ ways to arrange the members, the total combinations are $X \times Y$. Addition would only be used if these were mutually exclusive scenarios, but here, they all happen simultaneously to form one complete seating arrangement.

Q: Does the order of the families matter? Why is it $3!$ and not just $1$?

It absolutely matters! The families are distinct entities. Family 1 sitting at the start of the row is a completely different seating arrangement than Family 2 sitting at the start. Since we have 3 distinct blocks, we have $3!$ ways to order them. If the families were identical, we would divide by $3!$, but in this problem, they are unique groups.

Q: I keep getting confused about when to use factorial. Why is it $4!$ for the last family?

Factorial is simply the number of ways to arrange $n$ distinct items in a row. For the third family, there are 4 members. If you have 4 distinct people, they can be arranged in $4 \times 3 \times 2 \times 1 = 24$ ways, which is $4!$. Always count the number of items *inside* the block you are currently considering, and apply the factorial to that specific count.

Question

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?

Options:

(A) $2! 3! 4!$

(B) $(3!)^3 (4!)$

(C) $(3!)^2 (4!)$

(D) $3!(4!)^3$

Sigma Percentile · Accepted Answer

### Phase 1: The Power of the Block Method

Imagine you are an event manager. You have 10 people, but they are stubborn and refuse to sit apart from their families. We have Family 1 (3 members), Family 2 (3 members), and Family 3 (4 members).

If we tried to arrange these 10 people individually, we would be lost in a sea of constraints. This is where the **Block Method** comes to our rescue.

We treat each family as a single, unbreakable unit. By tying each family together with an invisible rope, we reduce 10 individuals into 3 distinct 'super-members' or 'blocks'. This transformation is the key to unlocking the problem.

### Phase 2: The External Dance

Now that we have our 3 blocks—let us call them $B_1$, $B_2$, and $B_3$—we need to arrange them in a row. Since these are 3 distinct families, the order in which the blocks appear matters.

The number of ways to arrange $n$ distinct objects is given by $n!$. Therefore, the number of ways to arrange our 3 family blocks is $3!$.

This accounts for all possible sequences, such as Family 1, then Family 3, then Family 2, and so on. We have successfully handled the 'macro' arrangement.

### Phase 3: The Internal Dance

But wait! We are not done. Within each block, the members are not statues; they can shuffle their seats. This is the 'micro' arrangement.

For Family 1, with 3 members, there are $3!$ ways to arrange them internally. For Family 2, also with 3 members, there are another $3!$ ways. For Family 3, with 4 members, there are $4!$ ways.

It is crucial to treat these as independent choices. The way Family 1 sits does not restrict how Family 3 sits; they are free to shuffle independently.

### Phase 4: The Grand Synthesis

Now, we apply the Fundamental Principle of Counting. To find the total number of ways to seat everyone, we multiply the number of ways to arrange the blocks by the number of ways to arrange the members within each block.

Mathematically, this is expressed as:

$$	ext{Total Ways} = (	ext{Arrangement of Blocks}) 	imes (	ext{Internal } F_1) 	imes (	ext{Internal } F_2) 	imes (	ext{Internal } F_3)$$

Substituting our values, we get:

$$	ext{Total Ways} = 3! 	imes 3! 	imes 3! 	imes 4!$$

This simplifies beautifully to $(3!)^3 	imes 4!$. Calculating the numerical value:

$$	ext{Total Ways} = 6 	imes 6 	imes 6 	imes 24 = 216 	imes 24 = 5184$$

Look at the elegance of this result! We have taken a chaotic scenario of 10 people and reduced it to a simple, structured product of factorials.

This is the beauty of combinatorics—it turns complexity into order. Keep this logic in your toolkit, and no permutation problem will ever intimidate you again. You have mastered the **Block Method**!

Animated Solution for Mathematics - Permutations and Combinations: 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?

Select Answer:

Visualized Solution

Visualizing the $3$ Families

The Block Method Concept

Arranging the $3$ Family Blocks

Internal Arrangement: $F_{1}$

Internal Arrangement: $F_{2}$

Internal Arrangement: $F_{3}$

Applying Multiplication Principle

Final Result: $(3!)^{3} \times 4!$

The Sigma Insight: Linear Permutations

Phase 1

Phase 2

Phase 3

Phase 4

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