#### Answer

a. $120$
b. $6$
c. $\displaystyle \frac{1}{20}$

#### Work Step by Step

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)=$\displaystyle \frac{6}{120}=\frac{1}{20}$