In how many ways 5 identical red balls and 7 distinct black balls can be arranged in a line such that no two red balls are adjacent
Question
In how many ways can 5 identical red balls and 7 distinct black balls be arranged in a line such that no two red balls are adjacent?
Solution
The problem can be solved using the concept of permutations and combinations.
Step 1: Arrange the 7 distinct black balls. They can be arranged in 7! ways.
Step 2: Now, consider the spaces between the black balls. There are 8 such spaces (including the ends).
Step 3: We need to place 5 identical red balls in these spaces such that no two red balls are adjacent. This is equivalent to selecting 5 spaces out of 8 to place the red balls. This can be done in 8C5 ways.
Step 4: Since the red balls are identical, there is only 1 way to arrange them in the selected spaces.
So, the total number of ways to arrange the balls is 7! * 8C5 * 1 = 7! * 56 = 2540160 ways.
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