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

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

 

Option: 1

3,250,741


Option: 2

1,280,458


Option: 3

7,054,320


Option: 4

2,000,456


Answers (1)

best_answer

To solve this problem, we can consider the blue balls as separators or dividers that create separate regions for the red balls. We need to find the number of ways to arrange the 11 red balls in these regions, ensuring that no two blue balls are adjacent.

Let's represent the blue balls as 'B' and the red balls as 'R'. We can create regions for the red balls by placing the blue balls in between them. The total number of blue balls is 9, and we need to place them in such a way that no two blue balls are adjacent. To achieve this, we can use the concept of stars and bars, where the blue balls act as bars and the red balls act as stars.

We have 11 red balls, and we need to divide them into 10 regions (to create spaces for the blue balls) by placing 9 blue balls. This can be represented as:

RRRRRRRRRR

We have 10 spaces between the red balls where we can place the blue balls. Let's represent these spaces with underscores ('_'):

R_R_R_R_R_R_R_R_R_R

Now, we need to select 9 spaces out of these 10 spaces to place the blue balls. We can choose the positions of the blue balls in "10 choose 9" ways:

\mathrm{C(10,9)=10}

Therefore, there are 10 ways to select the positions of the blue balls.

Once we have placed the blue balls, we are left with 11 regions (including the ends) where we can distribute the 11 red balls. Each region can have any number of red balls (including zero). We can represent the distribution of the red balls using numbers as follows:

R_1 R_2 R_3 R_4 R_5 R_6 R_7 R_8 R_9 R_10 R_11

We can use the concept of stars and bars again to find the number of ways to distribute the red balls among the regions. Since we have 11 red balls and 11 regions, we can represent this as placing 11 stars among 11 regions (including the ends). The number of ways to distribute the red balls is then given by:

\mathrm{C(11+11,11)=C(22,11)=705,432}

Therefore, there are 705,432 ways to distribute the red balls among the regions.

Finally, we multiply the number of ways to select the positions of the blue balls (10 ways) by the number of ways to distribute the red balls among the regions (705,432 ways) to get the total number of arrangements:

Total arrangements = \mathrm{10 \times 705,432=7,054,320}

Therefore, there are 7,054,320 ways to arrange the red balls, 9 blue balls, and 7 green balls such that no two blue balls are adjacent.


 

Posted by

vishal kumar

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