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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 at least one green ball

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 at least one green ball separates any two red balls?

 

Option: 1

320


Option: 2

100


Option: 3

220


Option: 4

425


Answers (1)

best_answer

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

Let's consider the possible arrangements step by step:

Step 1: Place the red balls

Since the red balls are identical, we don't need to consider their specific order. We have 11 red balls to arrange. We can think of them as distinct objects separated by green balls:

G R G R G R G R G R G R G

Step 2: Place the blue and green balls

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

To calculate the number of possible arrangements, we can consider the spaces between the red balls. We have 12 spaces (including the ends) where we can place the blue and green balls.

Using the concept of stars and bars, we need to distribute 9 blue balls and 7 green balls among the 12 spaces.

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

C(12,9)=12 ! /(9 ! \times(12-9) !)=12 ! /(9 ! \times 3 !)=(12 \times 11 \times 10) /(3 \times 2 \times 1)=220

Therefore, the number of ways the red balls can be arranged, given the conditions, is 220.

 

Posted by

Pankaj

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