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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 red balls are ad

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 red balls are adjacent?

 

Option: 1

7,30,250


Option: 2

5,10,280


Option: 3

6,10,080


Option: 4

923,780


Answers (1)

best_answer

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

Let's consider the red balls as separators or dividers to create regions for the blue and green balls. We need to find the number of ways to arrange the 9 blue balls and 7 green balls in these regions, ensuring that no two red balls are adjacent.

We have 9 blue balls and 7 green balls, so we need to create 10 regions for the blue balls and 8 regions for the green balls. We can use the concept of stars and bars, where the blue and green balls act as stars and the red balls act as bars.

We have a total of 18 balls (9 blue balls + 9 red balls) and we need to divide them into 10 regions (9 for the blue balls and 1 for the green balls) by placing 9 red balls. This can be represented as:

BBBBBBBBB

We have 10 spaces between the blue balls and one space at the end where we can place the green ball. Let's represent these spaces with underscores ('_'):

B_B_B_B_B_B_B_B_B_B_

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

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

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

Once we have placed the red balls, we are left with 10 regions (including the ends) for the blue balls and one region for the green ball. Each region can have any number of blue or green balls (including zero). We can represent the distribution of the blue and green balls using numbers as follows:

B_1 B_2 B_3 B_4 B_5 B_6 B_7 B_8 B_9 B_10 G

We can use the concept of stars and bars again to find the number of ways to distribute the blue and green balls among the regions. Since we have 9 blue balls and 1 green ball, we can represent this as placing 9 stars among 10 regions. The number of ways to distribute the blue and green balls is then given by:

\mathrm{C(9+10,9)=C(19,9)=92,378}

Therefore, there are 92,378 ways to distribute the blue and green balls among the regions.

Finally, we multiply the number of ways to select the positions of the red balls (10 ways) by the number of ways to distribute the blue and green balls among the regions (92,378 ways) to get the total number of arrangements:

\mathrm{\text { Total arrangements }=10 \times 92,378=923,780}

Therefore, there are 923,780 ways to arrange the blue balls, red balls, and green balls such that no two red balls are adjacent.

 

Posted by

Divya Prakash Singh

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