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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 at least one red ball s

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 at least one red ball separates any two blue balls?

 

Option: 1

30,250


Option: 2

19,280


Option: 3

47,080


Option: 4

18,564


Answers (1)

best_answer

To calculate the number of ways the blue balls can be arranged under the given conditions, where at least one red ball separates any two blue balls, we can use the concept of permutations.

We have 11 identical red balls, 9 identical blue balls, and 7 identical green balls. We need to ensure that at least one red ball is placed between any two blue balls.

Let's consider the possible arrangements step by step:

Step 1: Place the blue balls

Since the blue balls are identical, we don't need to consider their specific order. We have 9 blue balls to arrange. We can think of them as distinct objects separated by red balls:

R B R B R B R B R B R B R B R B R

Step 2: Place the red and green balls

Between each pair of adjacent blue balls, we can place any number of red or green balls as long as there is at least one red ball between any two blue balls.

To calculate the number of possible arrangements, we can consider the spaces between the blue balls. We have 18 spaces (including the ends) where we can place the red and green balls.

Using the concept of stars and bars, we need to distribute 11 red balls and 7 green balls among the 18 spaces.

The number of ways to distribute the red and green balls can be calculated using combinations:

C(18,11)=18 ! /(11 ! \times(18-11) !)-18 ! /(11 ! \times 7 !)-(18 \times 17 \times 16 \times 15 \times 14 \times 13 \times 12) /(7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1)=18,564

Therefore, the number of ways the blue balls can be arranged, given the conditions, is 18,564.

 

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Gaurav

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