5 girls and 3 boys are arranged randomly in a row. Find the probability that:
1) there is one boy on each end.
2) There is one girl on each end. Select
A) 3/28 & 5/14  B) 5/28 & 3/14  C) 5/14 & 3/28  D) 3/28 & 1/14  E) 1/28 & 5/14
Total # of arrangements is 8!;   1. The # of arrangements where there is one boy on each end is C^2/3∗2∗6! where C^2/3C32 is # of ways to select which 2 boys will be on that positions, *2 # of way to arrange them on that position (A------B and B------A) and 6! is # of arrangements of 6 people left;   P=C^2/3∗2∗6!/8!=3/28   2. The # of arrangements where there is one girl on each end is C^2/5∗2∗7!, where C^2/5 is # of ways to select which 2 girls will be on that positions, *2 # of way to arrange them on that position (3/28 &amp; 5/14------B and B------3/28 &amp; 5/14) and 6! is # of arrangements of 6 people left;   P=C^2/5∗2∗6!/8!=5/14   3/28 &amp; 5/14nswer: 3/28 &amp; 5/14.
<result>A</result>
5 girls and 3 boys are arranged randomly in a row. Find the probability that:
1) there is one boy on each end.
2) There is one girl on each end.