Question

The number of ways of forming a queue of 4 boys and 3 girls such that all the girls are not together, is:

• A. 5040
• B. 3050
• C. 3410
• D. 4320

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
We are asked to find the permutations of 7 people where 3 specific people (the girls) are NOT all standing together. The easiest approach is using the complement rule:
(Total ways to arrange all people) \(-\) (Ways to arrange them such that all 3 girls are together).
Step 2: Key Formula or Approach:
Total number of people = 4 boys + 3 girls = 7 people.
Total number of arrangements without any restriction = \(7!\).
Number of ways to arrange them together = \((n - k + 1)! \times k!\) where \(k\) is the size of the group.
Step 3: Detailed Explanation:
Let's calculate the total ways:
\[ 7! = 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 5040 \] Now, consider the case where all 3 girls are together. We can treat the 3 girls as a single unit or "block".
The new number of entities to arrange is 4 boys + 1 block of girls = 5 units.
These 5 units can be arranged in \(5!\) ways.
Within the block, the 3 girls can be arranged among themselves in \(3!\) ways.
Number of ways all 3 girls are together = \(5! \times 3! = 120 \times 6 = 720\).
Step 4: Final Answer:
The number of ways such that all 3 girls are NOT together is:
\[ \text{Total ways} - \text{Ways all girls together} = 5040 - 720 = 4320 \]