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There are 7 identical white balls and 3 identical black balls. The number of distinguishable arrangements in a row of all the balls, so that no two black balls are adjacent, is
(A) 120;
(B) 89(8!);
(C) 56;
(D) 42x5^4.

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you can arrange the 7 white balls in 7! ways as all balls are identical so it should be (7!/7!)

now for black balls you have 8 places. from 8 places you can choose any of the 3 places= 8C

and you can arrange them in 3! ways as all the 3 balls are identical so it should be (3!/3!)

hence solution should be (7!/7!)*8C3 *(3!/3!) = 56

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