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  • In how many ways can the red, blue, and green balls be arranged if there are 11 identical red balls, 9 identical blue balls, and 7 identical green balls such that no two

In how many ways can the red, blue, and green balls 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

20 ! / 11 !


Option: 2

20 ! / 18 !


Option: 3

18 ! / 11 !


Option: 4

20 ! / 10 \text { ! }


Answers (1)

best_answer

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

We have 11 identical red balls, 9 identical blue balls, and 7 identical green balls. To ensure that no two balls of the same color are adjacent, we need to place the balls in an alternating pattern.

We can start by arranging the red and blue balls. We have a total of 20 balls (11 red + 9 blue) to arrange in an alternating pattern. We can think of this as arranging 20 slots where we can place the red and blue balls, ensuring that no two balls of the same color are adjacent.

Using the concept of stars and bars, we have 20 slots and 2 types of balls (red and blue). We need to place the red and blue balls in these slots, ensuring that no two balls of the same color are adjacent.

Let's represent the arrangement as follows, where "R" represents a red ball, "B" represents a blue ball, and "|" represents a space:

R | B | R | B | R | B | R | B | R | B | R | B | R | B | R | B | R | B | R | B |

Now, we have 20 spaces (represented by the "|" symbols) where we can place the balls. We need to select positions for the 11 red balls and 9 blue balls.

Using the combination formula \mathrm{(n C r)} we can calculate the number of ways to choose positions for the red and blue balls:

\mathrm{\begin{aligned} & C(20,11) \times C(9,9)=(20 ! /(11 ! \times(20-11) !)) \times(9 ! /(9 ! \times(9-9) !)) \\ & =(20 ! /(11 ! \times 9 !)) \times(9 ! / 9 !) \\ & =(20 ! / 11 !) \times 1 \\ & =20 ! / 11 ! \end{aligned}}

Therefore, the number of ways the red, blue, and green balls can be arranged, given the conditions, is \mathrm{20 ! / 11 !}

 

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Riya

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