First slide

Permutations

Question

In a class of 10 students there are 3 girls students. The number of ways in which they can be arranged in a row, so that no two girls are consecutive is $k (8!)$ , then $k$ is equal to

Moderate

A

12
B

24
C

36
D

42

Solution

Seven boys can be arranged in 7! ways.
If $X B_{1} X B_{2} X B_{3} X B_{4} X B_{5} X B_{6} X B_{7} X$ s one such arrangement,
then 3 girls can be arranged at any of the 3 places marked with a X.
the number of arrangements is $(^{8} P_{3}) (7!) = (6 \times 7) (8!) = (42) (8!)$ Thus, $K = 42 .$

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