## Thinking Mathematically (6th Edition)

a. $120$ b. $6$ c. $\displaystyle \frac{1}{20}$
The order of arrival is of importance: permutations are involved. a. The total ways in which they can arrive is: ${}_{5}P_{5}=5!=5\times 4\times 3\times 2\times 1=120$ b. If the first and last events have been determined: $( M, {\_} , {\_} , {\_} , A),$ the middle three can arrive in ${}_{3}P_{3}=3!=6$ ways c. Let E be the described event. P(E)$=\displaystyle \frac{the\ number\ of\ ways\ the\ permutation\ can\ occur}{total\ number\ of\ possible\ permutations}$ P(E)=$\displaystyle \frac{6}{120}=\frac{1}{20}$