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

In how many ways can the red, blue, and 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 ball of a different color separates any two balls of the same color?

 

Option: 1

\mathrm{C(21,9) \times C(20,5)}


Option: 2

\mathrm{C(21,9) \times C(18,7)}


Option: 3

\mathrm{C(20,9) \times C(20,7)}


Option: 4

\mathrm{C(21,9) \times C(20,7)}


Answers (1)

best_answer

To calculate the number of ways the red, blue, and green balls can be arranged under the given conditions, where at least one ball of a different color separates any two balls of the same color, 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 ball of a different color separates any two balls of the same color.

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.

Step 2: Place the blue balls

Between each pair of adjacent red balls, we need to place at least one blue ball to ensure that a different color separates any two red balls. We have 9 identical blue balls to distribute among the spaces.

Using the concept of stars and bars, we have 12 spaces (including the ends) where we can place the blue balls. We need to distribute the 9 blue balls among these 12 spaces.

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

\mathrm{C(12+9,9)=C(21,9)}

Step 3: Place the green balls

Between each pair of adjacent balls (red or blue), we need to place at least one green ball to ensure that a different color separates any two balls of the same color. We have 7 identical green balls to distribute among the remaining spaces.

Using the concept of stars and bars, we have 13 spaces (including the ends) where we can place the green balls. We need to distribute the 7 green balls among these 13 spaces.

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

\mathrm{C(13+7,7)=C(20,7)}

Therefore, the number of ways the red, blue, and green balls can be arranged, given the conditions, is \mathrm{C(21,9) \times C(20,7)}

 

Posted by

HARSH KANKARIA

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