Permutation and CombinationMedium
Question
The number of ways, of forming a queue of 4 boys and 3 girls such that all the girls are not together, is:
Options
A.5040
B.3050
C.3410
D.4320
Solution
Sol. [Key Concept: Total minus all-girls-together: $7! - 5!\ 3!$.]
No. of ways to arrange 7 students $= 7$ !
Cases where all girls are together
$$B_{1}B_{2}B_{3}B_{4}\boxed{\left) G_{1}G_{2}G_{3} \right.\ } = 5!3! $$Ans. $= 7! - 5!3!$
$= 7! - 6$ !
$${= 6 \times 6! }{= 6 \times 720 }{= 4320 }$$
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