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    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.
    A 4600 Correct Answer Incorrect Answer
    B 4464 Correct Answer Incorrect Answer
    C 4320 Correct Answer Incorrect Answer
    D 5040 Correct Answer Incorrect Answer
    E 4840 Correct Answer Incorrect Answer

    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 ā€“ (24 Ɨ 24) = 4464 ways

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