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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 blue ball s

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 blue ball separates any two red balls?

 

Option: 1

30


Option: 2

55


Option: 3

25


Option: 4

17


Answers (1)

best_answer

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

We can treat the blue balls as dividers or separators between the red balls. Let's represent the arrangement using "R" for red balls and "|" for blue ball separators:

R | R | R | R | R | R | R | R | R | R | R |

Now, we need to place the 9 blue balls in the spaces between the red balls and at the ends of the arrangement. We have 11 spaces (represented by the "|" symbols) where we can place the blue balls.

Using the concept of stars and bars, we need to select 9 out of the 11 spaces to place the blue balls. We can calculate this using the combination formula (\mathrm{n C r):}

\mathrm{\begin{aligned} & C(11,9)=11 ! /(9 ! \times(11-9) !) \\ & =11 ! /(9 ! \times 2 !) \\ & =(11 \times 10) /(2 \times 1) \\ & =55 \end{aligned}}

 

 

 

 

Posted by

Sanket Gandhi

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