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

Answer:

$\frac{n!}{r!(n-r)!}$

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 $\binom{n}{r}$ 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,

$r! \times (n-r)!$ permutations.

So the actual amount is,

$\frac{r!}{(n-r)!}= \binom{n}{r}=\binom{n}{n-r}$