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Exercise 6.2 · Q5

Q.There are 5 vacant seats in a row. In how many ways can 3 men sit.

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Seating 3 men in 5 distinct vacant seats (order matters — different seats are different arrangements) gives 5P3=60^{5}P_3=60 ways.

[!FORMULA] nPr=n!(n−r)!^{n}P_{r}=\dfrac{n!}{(n-r)!} counts the number of ways to arrange rr distinct objects into rr of nn distinct positions, where nn is the number of available seats and rr is the number of men to be seated.

  1. Here n=5n=5 (vacant seats in a row) and r=3r=3 (men to seat). Since seats are distinguishable (position in the row matters) and the men are distinct, this is a permutation problem.

  2. Apply the formula: 5P3=5!(5−3)!=5!2!^{5}P_3=\dfrac{5!}{(5-3)!}=\dfrac{5!}{2!}.

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