Dr Bhishma Hazarika
Last Activity: 6 Years ago
Method I:
(a) RGRGRGRGR (5 red and 4 Green), As Identical balls this is one arragnement (1)
(b) ___G____G____G____G(Four places for 5 red balls. 2 red balls wil be placed in one place and remaining one for one), So in 4 way it can be done (4)
(c)G___G___G___G___ (fill up like option b in reverse order) again (4)
(d) G___G___G___G (Three places for 5 red balls. 2 balls will be placed in any two places and remaining one will be placed in remaining place), so 3 way it can be done (3)
So answer is 1+4+4+3=12
Method II:
Let take the arrangement ____R____R_____R_____R_____R____
So 4 green balls can be arranged in 6 places, so it is in 6 C4 way, but in some of the arrangement 3 red balls comes together.We can’t take those arrangement. those are GRRRGRGRG, GRGRGRRRG and GRGRRRGRG
So answer is 6 C4 -3=15-3=12
