I have $20$ blue balls and $25$ red balls. All of the $45$ balls are randomly placed in a row. The blue and red balls form a pair when there is a red ball left from a blue ball. If the first ball is a blue ball on the first spot and a red ball on the last spot, then it also counts as a pair. What is the probability that there are exactly $15$ pairs?
I tried dividing the problem into 4 categories:
b.....b ,
r.....r,
r.....b,
b....r,
With the first $3$ categories, you still need to make $15$ more pairs and with the last category you need to make only $14$ more pairs. Then I tried to find the number of possible ways to make exactly $15$ more pairs in the first three categories, but I got stuck on how to do that. I also know that the probability is equal to the probability of the row consisting of exactly $15$ groups of blue balls and $15$ groups of red balls.
To make $15$ groups of blue balls and $15$ groups of red balls I first need $14$ red balls to divide my $15$ groups of blue balls, then I need $14$ blue balls to divide my $15$ groups of red balls. Now there are $45 - 14 - 14 = 17$ balls left, $6$ blue balls and $11$ red balls. Then there are $C(17,6) \cdot C(17,11)$ possible combinations? I'm not sure if this is right.
There are a total of $$\binom{45}{20}$$ sequences that can be formed with $20$ blue and $25$ red balls.
A run is a sequence of one or more balls of the same color.
If there are $15$ transitions from red to blue, then we have the following possibilities:
The first ball is red, and there are $15$ runs of red balls and $15$ runs of blue balls: Let $r_i$ be the number of red balls in the $i$th run, where $1 \leq i \leq 15$. Then $$r_1 + r_2 + r_3 + \cdots + r_{15} = 25$$ is an equation in the positive integers.
A particular of the equation $$x_1 + x_2 + \cdots + x_k = n$$ in the positive integers corresponds to the placement of $k - 1$ addition signs in the $n - 1$ spaces between successive ones in a row of $n$ ones.
To illustrate, suppose that $k = 4$ and $n = 10$. Then we wish to place three addition signs in the nine spaces between successive ones in a row of ten ones. $$1 \square 1 \square 1 \square 1 \square 1 \square 1 \square 1 \square 1 \square 1 \square 1$$ If we pick the third, fifth, and seventh spaces, we obtain $$1 1 1 + 1 1 + 1 1 + 1 1 1$$ which corresponds to the solution $x_1 = 3$, $x_2 = 2$, $x_3 = 2$, $x_4 = 3$.
The number of solutions of the equation $x_1 + x_2 + x_3 + \cdots + x_k = n$ corresponds to the number of ways we can place $k - 1$ addition signs in a row of $n$ ones, which is $$\binom{n - 1}{k - 1}$$ which we must choose which $k - 1$ of the $n - 1$ spaces between successive ones in a row with $n$ ones will be filled with addition signs.
Therefore, there are $$\binom{25 - 1}{15 - 1}$$ ways to have $15$ runs of red balls.
Let $b_i$ be the number of blue balls in the $i$th run, where $1 \leq i \leq 15$. Then $$b_1 + b_2 + b_3 + \cdots + b_{15} = 20$$ is an equation in the positive integers with $$\binom{20 - 1}{15 - 1} = \binom{19}{14}$$ solutions.
Thus, there are $$\binom{24}{14}\binom{19}{14}$$ sequences beginning with a red ball in which there are $15$ runs of red balls and $15$ runs of blue balls.
The first ball is red, and there are $16$ runs of red balls and $15$ runs of blue balls: Let $r_i$ be the number of red balls in the $i$th run, where $1 \leq i \leq 16$. Then $$r_1 + r_2 + r_3 + \cdots + r_{16} = 25$$ is an equation in the positive integers with $$\binom{25 - 1}{16 - 1} = \binom{24}{15}$$ solutions.
Let $b_i$ be the number of blue balls in the $i$th run, where $1 \leq i \leq 15$. Then $$b_1 + b_2 + b_3 + \cdots + b_{15} = 20$$ is an equation in the positive integers with $$\binom{20 - 1}{15 - 1} = \binom{19}{14}$$ solutions.
Thus, there are $$\binom{24}{15}\binom{19}{14}$$ sequences beginning with a red ball in which there are $16$ runs of red balls and $15$ runs of blue balls.
The first ball is blue, and there are $16$ runs of blue balls and $15$ runs of red balls: Let $b_i$ be the number of blue balls in the $i$th run, where $1 \leq i \leq 16$. Then $$b_1 + b_2 + b_3 + \cdots + b_{16} = 20$$ is an equation in the positive integers with $$\binom{20 - 1}{16 - 1} = \binom{19}{15}$$ solutions.
Let $r_i$ be the number of red balls in the $i$th run, where $1 \leq i \leq 15$. Then $$r_1 + r_2 + r_3 + \cdots + r_{15} = 25$$ is an equation in the positive integers with $$\binom{25 - 1}{15 - 1} = \binom{24}{14}$$ solutions.
Thus, there are $$\binom{19}{15}\binom{24}{14}$$ sequences beginning with a blue ball in which there are $16$ runs of blue balls and $15$ runs of red balls.
The first ball is blue, and there are $15$ runs of blue balls and $15$ runs of red balls: Let $b_i$ be the number of blue balls in the $i$th run, where $1 \leq i \leq 15$. Then $$b_1 + b_2 + b_3 + \cdots + b_{15} = 20$$ is an equation in the positive integers with $$\binom{20 - 1}{15 - 1} = \binom{19}{14}$$ solutions.
Let $r_i$ be the number of red balls in the $i$th run, where $1 \leq i \leq 15$. Then $$r_1 + r_2 + r_3 + \cdots + r_{15} = 25$$ is an equation in the positive integers with $$\binom{25 - 1}{15 - 1} = \binom{24}{14}$$ solutions.
Thus, there are $$\binom{19}{14}\binom{24}{14}$$ sequences beginning with a blue ball in which there are $15$ runs of blue balls and $15$ runs of red balls.
Total: Since these four cases are mutually exclusive and exhaustive, the number of favorable cases is $$\binom{24}{14}\binom{19}{14} + \binom{24}{15}\binom{19}{14} + \binom{19}{15}\binom{24}{14} + \binom{19}{14}\binom{24}{14}$$
Hence, the desired probability is $$\frac{\dbinom{24}{14}\dbinom{19}{14} + \dbinom{24}{15}\dbinom{19}{14} + \dbinom{19}{15}\dbinom{24}{14} + \dbinom{19}{14}\dbinom{24}{14}}{\dbinom{45}{20}}$$