The number of binary sequences of length 300 made up of 50 ones and 250 zeros so that between Every two ones has at least one zero.

Question

I tried a scenario where the number 0 represents a red ball and the number 10 represents a blue ball. I have a collection of 200 red balls, each occupying one spot, and 50 blue balls, each occupying two spots. The objective is to distribute these 250 balls across a total of 300 places.

I feel like I'm not in the right direction at all and would appreciate some help

Answer 1

Since I am going to take a different approach, I will use different colors. Suppose we have $50$ green and $250$ yellow balls. We wish to arrange the balls in a row so that no two of the green balls are adjacent. First, we will set aside $49$ of the yellow balls to be used later. That leaves $50$ green balls and $201$ yellow balls. They can be arranged in $\binom{251}{50}$ ways. Next, place a yellow ball to the immediate right of each of the first $49$ green balls to ensure that no two of the green balls are adjacent. Place a $1$ on each green ball and a $0$ on each yellow ball. The resulting binary sequence has the property that there is at least one $0$ between each two $1$s. Thus, the number of admissible arrangements is $\binom{251}{50}$.

Answer 2

• Arrange the $250$ zeroes in a row

• In the $251$ interstices (including the ends) between the zeroes, insert the $1's$ in $\binom{251}{50}$ ways