Question

How many different bit strings can be transmitted if the string must begin with a 1 bit, must include three additional 1 bits (so that a total of four 1 bits is sent), must include a total of 12 0 bits, and must have at least two 0 bits following each 1 bit?

Short answer

There are 12,565,671,261 ways to assign scores to the problems

Step-by-step solution

Step 1: Use the formula for integer

Formula for integer

\(C\left( {n + r - 1,r} \right) = \frac{{(n + r - 1)!}}{{n!r!}}\)

n is number of bits

r is number of spots

Step 2: Solution of the number of strings

Let’s, applied n and r values of inequality in integer formula

There are 3 more 1 bits followed by two zeros we must consider and the possible combinations

Since there are 13 bits to consider and 9 of those spots need to be arranged in the manner of a 1 bit followed by two 0 bits would the total possible number of combinations be

You have 3 elements of the form 100 and 4 zeros to place in 13 positions as you say (we ignore the first three because they are indeed fixed).

So, by considering the element 100 as a unity, you actually have 3 (compounded) elements and 4 zeros to place in 7 positions.

Ex:

\(\begin{array}{l}C\left( {n + r - 1,r} \right) = \frac{{(n + r - 1)!}}{{n!r!}}\\ = \frac{{7!}}{{3!4!}}\\ = 35\end{array}\)