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$6$ blue $7$ green and $10$ white balls are arranged in rows such that every blue ball is between and green and white ball. Moreover a white ball and a green ball must not be next to each other. Number of such arrangements is 2 0 comment dev

## Case 1

Consider the arrangement of the balls as follows (G+ denotes $1$ or more green balls, B denotes $1$ blue ball, W+ denotes $1$ or more white balls) First place the $6$ blue balls. Then the $7$ green balls can be arranged into the $4$ positions in $6\text{C}3$ ways.

(We got $6\text{C}3$ above because the arrangement is similar to distribution of $7$ identical balls into $4$ distinct boxes where each box must contain at least $1$ ball. Such distributions are explained in detail here. see formula $7$)

Then the $10$ white balls can be arranged in $3$ places in $9\text{C}2$ ways.

Total ways $=6\text{C}3×9\text{C}2=720$

## Case 2

Consider the arrangement of the balls as follows. Like the first case, place the $6$ blue balls. Then the $7$ green balls can be arranged into the $3$ positions in $6\text{C}2$ ways. Then the $10$ white balls can be arranged in $4$ places in $9\text{C}3$ ways.

Total ways $=6\text{C}2×9\text{C}3=1260$

Required number of ways $=720+1260=1980$ 0 0 comment

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