Question

The number of ways 4 boys and 3 girls are to be arranged in a row so that all 3 girls are not together, is equal to:

• A. 2320
• B. 4320
• C. 4920
• D. 1440

Hint:

Note the wording: "all 3 girls are not together" means you only subtract the case where the block of 3 is intact. It still allows for 2 girls to be together. If the question said "no two girls are together," you would use the Gap Method instead.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
We need to find the number of arrangements of 7 people (4 boys and 3 girls) such that all three girls are not together.
This is best solved using the complement method:
$\text{Required Ways} = \text{Total Ways} - \text{Ways where all 3 girls are together}$
Step 2: Key Formula or Approach:
1. Total arrangements of $n$ distinct objects is $n!$.
2. String method: To keep objects together, tie them as one unit.
Step 3: Detailed Explanation:
Total number of arrangements of 4 boys and 3 girls (7 people total):
\[ \text{Total} = 7! = 5040 \]
Now, consider the case where all 3 girls ($G_1, G_2, G_3$) are together.
Treat the 3 girls as one single unit (X).
Now we have 4 boys ($B_1, B_2, B_3, B_4$) and 1 unit (X), making 5 units in total.
Number of ways to arrange these 5 units:
\[ 5! = 120 \]
The 3 girls within the unit (X) can be arranged among themselves in $3!$ ways.
\[ \text{Girls together} = 5! \times 3! = 120 \times 6 = 720 \]
Therefore, the number of ways where all 3 girls are not together:
\[ \text{Required ways} = 5040 - 720 = 4320 \]
Step 4: Final Answer:
The number of ways is 4320.