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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 red 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 red ball separates any two green balls?

 

Option: 1

\mathrm{\left({ }^{26} C_{11}\right) \times\left({ }^{26} C_9\right)}


Option: 2

\mathrm{\left({ }^{28} C_{11}\right) \times\left({ }^{26} C_9\right)}


Option: 3

\mathrm{\left({ }^{28} C_{10}\right) \times\left({ }^{26} C_9\right)}


Option: 4

\mathrm{\left({ }^{28} C_{11}\right) \times\left({ }^{20} C_9\right)}


Answers (1)

best_answer

To calculate the number of ways the green balls can be arranged under the given conditions, where at least one red 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 red 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 red balls:

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

Step 2: Place the red and blue balls

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

To calculate the number of possible arrangements, we can consider the spaces between the green balls and the ends. We have 17 spaces where we can place the red and blue balls.

Using the concept of stars and bars, we need to distribute 11 red balls and 9 blue balls among the 17 spaces.

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

\mathrm{C(17+11,11) \times C(17+9,9)=\left({ }^{28} C_{11}\right) \times\left({ }^{26} C_9\right)}

Therefore, the number of ways the green balls can be arranged, given the conditions, is \mathrm{\left({ }^{28} C_{11}\right) \times\left({ }^{26} C_9\right)}

 

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Anam Khan

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