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.

Class: 10

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{10! \times}^{}^{9} C_{2}

{10! \times}^{}^{7} C_{2}

{10! \times}^{}^{8} C_{2}

{11! \times}^{}^{9} C_{2}

answer is A.

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Detailed Solution

Given, We have 10 red balls of different shades and 9 green balls, we have to select two balls such that not two green balls are together.
Number of ways for 10 red balls for different shades= 10!
Number of ways for Green balls =

{^{}}^{9} C_{2}

Number of ways=

{10! \times}^{}^{9} C_{2}

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