Question
Seven balls out of which 4 are red and the rest are green
are to be arranged in a row. Find the number of arrangements in which all four red balls are not placed together.Solution
Total number of ways of arrangement = 7! = 5040 ways
If we consider 4 red balls as one item, number of arrangements among themselves = 4! = 24 ways
Number of arrangements of remaining 4 items = 4! = 24 ways
Therefore, required number of ways = 5040 – (24 × 24) = 4464 ways
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