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

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Aicharya · Accepted Answer

We will use Gap method to solve this.

Firstly, $7$ White balls can be arranged in $7!$ ways. But Since all balls are identical so it should be $\frac{7!}{7!}$ ways.

Now, A black ball can be between these white balls.

So there are exactly $8$ slots to place $3$ black balls i.e., $__W__W__W__W__W__W__W__$

This can be done in ${}^{8}{C_{3}}$ ways.

Hence, the total number of ways is $\frac{7!}{7!} \times {}^{8}{C_{3} = 56}$ ways.

There are 7 identical white balls and 3 identical black balls. The number of distinguishable arrangements in…

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