Question

How many different bit strings can be formed using six 1s and eight 0s?

Short answer

Therefore, there are\(3003\) different strings formed.

Step-by-step solution

Step 1: Assumptions and substitution

The string contains six ones and eight zeros, thus fourteen bits in total.

\(\begin{array}{l}n = 6 + 8 = 14\\k = 2\end{array}\)

\(\begin{array}{l}{n_1} = 6\\{n_2} = 8\end{array}\)

Step 2: Further simplification

Distributing n distinguishable objects into k distinguishable objects are place in box i (\(i = 1,2,3,4,5\)) can be done in

\(\begin{array}{l}\frac{{n!}}{{{n_1}!{n_2}!......{n_k}!}} = \frac{{14!}}{{6!8!}}\\ = 3003\end{array}\)

Therefore, there are\(3003\) different strings formed.