Challenging Permutation question 4

In how many ways can 9 balls of which 4 are red, 4 are white and 1 black be arranged in a line so that no red ball is next to the black?

Answer: 175

Solution

Case 1: 1st ball is black in the whole row of 9 balls

Then the 2nd ball must be white. The rest of the 4 red and 3 white balls can be arranged in Tex2Img_1404380765= 35 ways

Case 2: last ball is black

Then the 2nd last ball must be white. The rest of the 4 red and 3 white balls can be arranged in Tex2Img_1404380765 = 35 ways

Case 3: Black ball is not the 1st or last ball

Then the black ball must be between 2 white balls. Consider WBW as one unit. Together with 4 red balls and 2 white balls, these can arranged in Tex2Img_1404380967 = 105 ways

Using the addition principle, total number of ways = 35+35+105= 175

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