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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 green ball

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 green ball separates any two blue balls?

 

Option: 1

352,716


Option: 2

810,280


Option: 3

510,780


Option: 4

356,000


Answers (1)

best_answer

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

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

Step 2: Place the red balls

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

To calculate the number of possible arrangements, we can consider the spaces between the blue balls and the ends. We have 10 spaces where we can place the red balls.

Using the concept of stars and bars, we need to distribute 11 red balls among the 10 spaces.

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

\mathrm{\begin{aligned} & C(10+11,11)=C(21,11)=21 ! /(11 ! *(21-11) !)=21 ! /(11 ! * 10 !)=(21 * 20 * 19 * 18 * 17 * 16 * \\ & 15 * 14 * 13 * 12) /(10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1)=352,716 \end{aligned}}

Therefore, the number of ways the blue balls can be arranged, given the conditions, is 352,716.


 

Posted by

Sanket Gandhi

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