In how many ways can $3$ identical red balls, $3$ identical green balls, and $3$ identical blue balls be arranged in a $3\times 3$ grid, such that each row and each column of the grid contains $1$ ball of each color

GIVEN: [tex]3[/tex] identical red balls, [tex]3[/tex] identical green balls, and [tex]3[/tex] identical blue balls be arranged in a [tex]3\times3[/tex] grid.

TO FIND: In how many ways can balls be arranged such that each row and each column contains [tex]1[/tex] ball of each color.

SOLUTION:

As each column and row must contain a red,a green and a blue ball.

Consider the image attached

each column contains a green,a blue and a red ball

To find more number of ways of arrangement, columns can be arranged differently such that each row will still contain each of three ball.

Now,

Total number of ways of arranging [tex]3[/tex] columns [tex]=3![/tex]

[tex]=3\times2\times1[/tex]

[tex]=6[/tex]

Hence total number of ways such that each row and each column of the grid contains [tex]1[/tex] ball of each color is [tex]6[/tex]