Answer
$3003$
Work Step by Step
We want to count the number of bit strings that consist of eight 1's and six 0's. This is equivalent to counting the number of permutations of eight 1's (indistinguishable from each other) and six 0's (indistinguishable from each other). By Theorem 3 on page 428, the total number of ways to do this is:
$\frac{14!}{8!6!}$
$=\frac{1*2*3*4*5*6*7*8*9*10*11*12*13*14}{1*2*3*4*5*6*7*8*1*2*3*4*5*6}$
$=\frac{9*10*11*12*13*14}{1*2*3*4*5*6}$
$=3003$
