The number of ways of arranging m white and n black balls in a row `(mgtn)` so that no two black are placed together, is

Options

A. `""^(m+1)C_(n)xxn!m!`
B. `""^(m)C _(n)`
C. `""^(m+1)C _(n)`
D. `m!n!`

Explanation

Since `mgtn` and we have to find the total number of arrangements of m+n balls so that no two black balls are together. So we first arrange m white balls in a row. This can be done in only one way as all white balls are identical.
Now, we have (m+1) places between m white balls and we want that no two black balls are together. So, we select n places out of (m+1) and then arrange them in these places. This can be done in
`""^(m+1)C _(n)xx1=""^(m+1)C_(n)` ways.
Hence, the required number of ways `""^(m+1)C _(n)`