Question

There are 10 red balls of different shades and 9 green balls of the same shades. The number of ways of arrangement such that no two green balls are together.

Answer 1

Hint:
first, we will arrange all the red balls. Then we will try to arrange green balls between every two red balls. Now since we are doing these two events successively we will multiply the number of possibilities of both events to get the final number of arrangements.

Complete step by step answer:
Now first let us just consider 10 red balls.
All these red balls are non-identical since they are of different shades
Hence we can arrange them at 10! Ways.
Now we can place a green ball either between two red balls or in the start or in the end.
By doing so we also confirm the condition that no two green balls are together.
Now there are 9 spaces in between red balls + 1 space at the start and + 1 in the end
Hence we have a total 11 places in which we can place green balls
But we have just 9 green balls
Hence Number of ways to arrange 9 green balls in 11 places is given by $^{11}{{C}_{9}}$ .
Since the two events are successive we multiply the number of arrangements to get the total number of arrangements
Hence total arrangements = $10{{!}^{11}}{{C}_{9}}$
So, the correct answer is “Option b”.

Note:
 Now we have all the green balls as identical hence we need not take care of arranging them after placing.