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  • Which of the following options represents the number of ways the red and blue balls can be arranged, if there are 11 identical red balls, 9 identical blue balls, and 7 id

Which of the following options represents the number of ways the red and blue balls can be arranged, if there are 11 identical red balls, 9 identical blue balls, and 7 identical green balls such that no two balls of the same color are adjacent? 

 

Option: 1

11 ! \times 9 ! \times 7 !


Option: 2

20 ! \times 21 !


Option: 3

18 ! \times 17 !


Option: 4

19 ! \times 20 !


Answers (1)

best_answer

To calculate the number of ways the red and blue balls can be arranged under the given conditions, we can use the concept of permutations.

Let's first consider the arrangement of red and blue balls without any restrictions. We have a total of 20 balls (11 red + 9 blue) to arrange.

The number of ways to arrange these balls without any restrictions can be calculated using the formula for permutations of indistinguishable objects:

\mathrm{P(20)=20 !}

Now, we need to subtract the arrangements where two balls of the same color are adjacent. We can treat the two balls of the same color as a single unit, so we have 2 units for the red balls and 2 units for the blue balls.

Now, we have 7 units (3 red-blue pairs + 1 green) and 14 spaces (20 total spaces - 6 occupied spaces by units). We need to arrange these 7 units and 14 spaces.

Using the concept of stars and bars, we can calculate the number of ways to arrange the units and spaces:

\mathrm{P(7+14)=P(21)=21 !}

Finally, we multiply the number of ways to arrange the red and blue balls by the number of ways to arrange the units and spaces:

\mathrm{P(20) \times P(21)=20 ! \times 21 !}

Therefore, the number of ways the red and blue balls can be arranged under the given conditions is

\mathrm{20 ! \times 21 !}

Posted by

Ritika Harsh

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