The sum of \(3\frac{7}{8}\) and \(1\frac{1}{3}\) is less than the difference between \(\frac{1}{8}\) and \(1\frac{2}{3}\) by:
A
3\(\frac{2}{3}\)
B
5\(\frac{1}{4}\)
C
6\(\frac{1}{2}\)
D
8
E
8\(\frac{1}{8}\)