Question

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 1

`(n!)/(r!(n-r)!)`

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
`\binom{n}{r}` 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
`r!(n-r)!` permutations. Since r black balls can be permuted in r! ways and white balls in (n-r)! different orders.

So basically it yields,

`r! * (n-r)!` permutations.

So the actual amount is,

`(r!)/((n-r)!)= \binom{n}{r}=\binom{n}{n-r}`