There are (n r) different linear arrangements of n balls of which r are black and n-r are white. give a combinatorial explanation of this fact

Answer:[tex]\frac{n!}{r!(n-r)!}[/tex]Step-by-step explanation:Combinatorial explanation: The n balls have to be arranged in n positions and the only distinction is where are the black and where white balls are. We can choose the position of black balls in [tex]\binom{n}{r}[/tex] ways, therefore, white ones are on the remaining positions. Using binomial we can have explanation written below: The balls can be arranged in n! possible permutations. To be precise one particular arrangement includes [tex]r!(n-r)![/tex] permutations. Since r black balls can be permuted in r! ways and white balls in (n-r)! different orders. So basically it yields, [tex]r! \times (n-r)![/tex] permutations. So the actual amount is, [tex]\frac{r!}{(n-r)!}= \binom{n}{r}=\binom{n}{n-r}[/tex]