Q.1.4 - A First Course in Probability

Question

There are$\left(\begin{array}{c}n\\ r\end{array}\right)$ different linear arrangements of $n$balls that $r$ are black and $n - r$ are white. Give a combinatorial explanation of this fact.

Step-by-Step Solution

Answer

Use the binomials.

Or consider the permutations of $n$ balls, and how many of them are the same by color.

Step 1 Explanation.

Possible orders of $r$ black, and $n-r$ white balls

The first solution - using combinations

The $n$ balls have to be arranged in $n$ positions, and the only distinction is where are the black, and where white balls.

We can choose the position of the black balls in $\left(\begin{array}{l}n\\ r\end{array}\right)$ ways, the white ones are in the remaining positions.

Step 2 Explanation.

The second solution - deriving the binomials.

The balls can be arranged in $n!$ possible permutations. But that counts as separate some cases that are the same in the color of the balls.

More precisely, one particular color arrangement includes $r!·(n-r)$permutations. Because the $r$ black balls can be permuted in $r!$ and the white balls in$n-r!$ different orders. By the basic principle, this yields $r!·(n-r)!$ permutations.

So the actual count is:

$\frac{n!}{r!·(n-r)!}=\left(\begin{array}{l}n\\ r\end{array}\right)=\left(\begin{array}{c}n\\ n-r\end{array}\right)$