Question 6 of 278MediumMCQ
The number of ways, of forming a queue of 4 boys and 3 girls such that all the girls are not together, is:
Solutions · 1
Official solution
We have a total of 4 boys and 3 girls.
Total number of children = 4+3=7.
The total number of ways to arrange 7 children in a queue is given by the permutation of 7 distinct objects taken all at a time, which is 7!.
7!=7×6×5×4×3×2×1=5040
Number of arrangements where all the girls are together
To find this, let us treat all 3 girls as a single unit or block, since they must remain together.
Now, we have 4 boys and 1 unit of girls. This gives us a total of 4+1=5 entities to arrange in the queue.
These 5 entities can be arranged among themselves in 5! ways.
Additionally, the 3 girls within their single unit can rearrange themselves in 3! ways.
By the Fundamental Principle of Counting (Multiplication Principle), the total number of ways in which all the girls are together is:
5!×3!
We know that 5!=120 and 3!=6, so:
120×6=720
Number of arrangements where all the girls are NOT together
To find the required number of ways, we subtract the arrangements where all girls are together from the total possible arrangements.
=−
=5040−720=4320
Therefore, the number of ways of forming a queue such that all the girls are not together is 4320.
The correct choice is Option D.
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