A bag contains 5 red, 4 black balls find the
number of arrangement such that
i) no two balls are of same colour together
ii) balls of same colour together

No. of red balls = 5
No. of black balls = 4
Total no, of balls = 5 + 4 = 9
They can be arranged in a row in 9! ways

(i) No. two balls of the same colour come together
First we arrange 4 black balls in a row in 4!
ways we get 5 gaps are formed among them as
shown below – B – B – B – B –
Now the 5 red balls can be filled with there 5
gaps in 5! ways
$\therefore$ Req. no. of arrangements = 4! × 5!
(ii) The balls of same colours come together:-
Treat the 5 red balls as one unit and 4 black as
1 unit
$\therefore$ 1 unit +1 unit =2 units
Then 2 units are arranged in a row in 2! ways
and 5 red balls are arranged among themselves
in 5! ways and 4 blacks are arranged among
them selves in 4! ways
$\therefore$Req. no. of arrangements = 2! × 5! × 4!