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In how many ways can 6 green toys and 6 red toys be arranged, such that 2 particular red toys are never together whereas 2 particular green toys are always together?


11! × 2!


9! × 90


4 × 10!


18 × 10!

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Option(D) is correct

Considering two green toys that are to be together as one unit.

We can arrange the 6 green toys and the remaining 4 red toys (excluding the 2 who are not to be together) is:

$= 9!×2!×{^{10}C_2}×2!$

$= \textbf{18× 10!}$

(4) Comment(s)


Considering 6 green toys as G1,G2,G3,G4,G5,G6 and 6 red toys as R1,R2,R3,R4,R5,R6. Assuming that two green toys that has to be always together as G1 and G2 and two red toys that has to be always separate as R1 and R2. Since G1 and G2 are always together let's consider it as one. Now except the constrains placed on R1 and R2 all the other toys can be placed in any order. Considering one random arrangement -


The toys R1 and R2 can be placed in any of the 10 vacant positions above. So the number of ways R1 and R2 can be placed is $10C_2*2!$.

The currently filled 9 positions can be arranged in 9! ways and G1 and G2 in 2! ways. Therefore total ways of arranging is $10C2*2!*9!*2!=18*10!$


U explained it better ...Thank You :)

Tejas Patil

answer is right, explanation is not upto the mark


can u please explain.Not able to understand.