In how many ways can 4 identical red balls, 6 identical green balls and 5 identical blue balls be arranged in a row such that all the blue balls are together and no two red balls are together?

This problem can be solved using the concept of permutations and combinations. Here are the steps:

1. First, consider the condition that all blue balls are together. We can treat the 5 identical blue balls as one single entity. So, we now have 4 red balls, 6 green balls and 1 "blue" entity.

2. The second condition is that no two red balls are together. To ensure this, we can place the red balls in the gaps between the green balls and the "blue" entity. There are 8 such gaps (7 between the green balls and 1 at the end of the "blue" entity).

3. The number of ways to place the 4 red balls in these 8 gaps is given by the combination formula C(n, r) = n! / [(n-r)!r!], where n is the total number of items, and r is the number of items to choose. So, C(8, 4) = 8! / [(8-4)!4!] = 70.

4. Now, we have 8 entities (4 red balls, 3 green balls and 1 "blue" entity) which can be arranged in 8! ways.

5. However, we have to divide by 4! for the red balls and 3! for the green balls because these balls are identical.

6. So, the total number of ways is 70 * (8! / (4! * 3!)) = 70 * 560 = 39200.

So, there are 39200 ways to arrange the balls according to the given conditions.

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To solve this problem, we need to use the principles of permutations and combinations.

Step 1: Treat the 5 identical blue balls as a single entity. So, we have 4 red balls, 6 green balls and 1 "blue" entity to arrange.

Step 2: Since no two red balls can be together, we can place them in the gaps between the green balls and the "blue" entity. We have 7 such gaps (6 between the green balls and 1 at each end). So, the red balls can be arranged in these gaps in 7P4 ways.

Step 3: Now, we have 6 green balls and 1 "blue" entity to arrange. They can be arranged in 7!/(6!*1!) ways.

Step 4: Within the "blue" entity, the 5 blue balls can be arranged in 5! ways.

So, the total number of ways = 7P4 * 7!/(6!*1!) * 5! = 840 * 7 * 120 = 705600 ways.

This problem can be solved using the concept of permutations and combinations. Here are the steps:

1. First, consider the condition that all blue balls must be together. We can treat the 5 identical blue balls as a single entity. So, we now have 4 red balls, 6 green balls, and 1 "blue" entity.

2. The condition that no two red balls are together implies that each red ball must be separated by at least one green ball or the "blue" entity. We can visualize this as placing the 4 red balls in the gaps between the green balls and the "blue" entity.

   We have 7 gaps where we can place the red balls (6 gaps between the green balls and 1 gap on either side of the "blue" entity).

3. The number of ways to place the 4 red balls in the 7 gaps is given by the combination formula C(n, r) = n! / [r!(n-r)!], where n is the total number of items, r is the number of items to choose, and "!" denotes factorial.

   So, the number of ways to place the red balls is C(7, 4) = 7! / [4!(7-4)!] = 35.

4. Once the red balls are placed, the green balls and the "blue" entity can be arranged in any order. The number of ways to arrange these 7 entities (6 green balls and 1 "blue" entity) is given by the permutation formula P(n, r) = n! / (n-r)!, where n is the total number of items and r is the number of items to arrange.

   Since we are arranging all 7 entities, r = n, so the formula simplifies to P(n, n) = n!.

   So, the number of ways to arrange the green balls and the "blue" entity is 7! = 5040.

5. Therefore, the total number of ways to arrange the balls according to the given conditions