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

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

 

Option: 1

792 \times 17 ! /(9 ! \times 8 !)


Option: 2

72 \times 17 ! /(9 ! \times 8 !)


Option: 3

792 \times 17 ! /(2 ! \times 8 !)


Option: 4

792 \times 19 ! /(9 ! \times 8 !)


Answers (1)

best_answer

To find the number of ways to arrange the blue balls, red balls, and green balls such that no two green balls are adjacent, we can use a similar approach as before.

We have 11 identical red balls, 9 identical blue balls, and 7 identical green balls. Let's represent the red balls as 'R', the blue balls as 'B', and the green balls as 'G'.

First, let's consider the red and blue balls together as separators or dividers to create regions for the green balls. We need to find the number of ways to arrange the 7 green balls in these regions, ensuring that no two green balls are adjacent.

We have 11 red balls and 9 blue balls, so we need to create 12 regions for the green balls. We can use the concept of stars and bars, where the green balls act as stars and the red and blue balls act as bars.

We have a total of 20 balls (11 red balls + 9 blue balls) and we need to divide them into 12 regions by placing 7 green balls. This can be represented as:

RRRRRRRRRRBBB

Now, we need to select 7 positions out of these 12 positions to place the green balls. We can choose the positions of the green balls in "12 choose 7" ways:

\mathrm{C(12,7)=792}

Therefore, there are 792 ways to select the positions of the green balls.

Once we have placed the green balls, we are left with 8 regions for the blue balls. Each region can have any number of blue balls (including zero). We can represent the distribution of the blue balls using numbers as follows:

B_1 B_2 B_3 B_4 B_5 B_6 B_7 B_8

 

Posted by

rishi.raj

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