six stand - up comics, a, b, c, d, e, and f, are to perform on a single evening at a comedy club. the order…

Explanation:

Step1: Calculate total number of arrangements

The number of permutations of (n) distinct objects is (n!). Here (n = 6), so the total number of ways the 6 comics can perform is (n!=6!=6\times5\times4\times3\times2\times 1=720).

Step2: Solve part a

Fix comic D in the fifth - position. Then the remaining 5 comics can be arranged in (5! = 5\times4\times3\times2\times1 = 120) ways. The probability (P(D\text{ fifth})=\frac{5!}{6!}=\frac{120}{720}=\frac{1}{6}).

Step3: Solve part b

Fix comic F in the second - position and comic E in the fifth - position. Then the remaining 4 comics can be arranged in (4! = 4\times3\times2\times1=24) ways. The probability (P(F\text{ second and }E\text{ fifth})=\frac{4!}{6!}=\frac{24}{720}=\frac{1}{30}).

Step4: Solve part c

There is only 1 way to arrange the comics in the order D, F, E, B, C, A. So the probability (P(D,F,E,B,C,A)=\frac{1}{6!}=\frac{1}{720}).

Step5: Solve part d

The number of ways comic C can perform first is (5!) (since the remaining 5 comics can be arranged in the remaining 5 positions). Similarly, the number of ways comic A can perform first is (5!). The number of ways C or A can perform first is (2\times5!). The probability (P(C\text{ or }A\text{ first})=\frac{2\times5!}{6!}=\frac{2\times120}{720}=\frac{1}{3}).

Answer:

a. (\frac{1}{6}) b. (\frac{1}{30}) c. (\frac{1}{720}) d. (\frac{1}{3})