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  • In how many ways can the green balls be arranged if there are 11 identical red balls, 9 identical blue balls, and 7 identical green balls such that at least one blue ball

In how many ways can the green balls be arranged if there are 11 identical red balls, 9 identical blue balls, and 7 identical green balls such that at least one blue ball separates any two green balls?

 

Option: 1

\mathrm{C(20,9) \times C(29,11)}


Option: 2

\mathrm{C(27,9) \times C(20,11)}


Option: 3

\mathrm{C(27,9) \times C(29,11)}


Option: 4

\mathrm{C(27,8) \times C(29,11)}


Answers (1)

best_answer

To calculate the number of ways the green balls can be arranged under the given conditions, where at least one blue ball separates any two green 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 blue ball is placed between any two green balls.

Let's consider the possible arrangements step by step:

Step 1: Place the green balls

Since the green balls are identical, we don't need to consider their specific order. We have 7 green balls to arrange. We can think of them as distinct objects separated by blue balls:

B G B G B G B G B G B G B G B G B

Step 2: Place the blue and red balls

Between each pair of adjacent green balls, we can place any number of blue or red balls.

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

Using the concept of stars and bars, we need to distribute 9 blue balls and 11 red balls among the 18 spaces.

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

\mathrm{C(18+9,9) \times C(18+11,11)=C(27,9) \times C(29,11)}

Therefore, the number of ways the green balls can be arranged, given the conditions, is \mathrm{C(27,9) \times C(29,11)}

 

 

Posted by

Pankaj Sanodiya

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