Question

Number of ways in which $7$ green bottles and $8$ blue bottles can be arranged in a row if exactly one pair of green bottles is side by side.

(Assume all bottles to be alike except color)

Answer 1

Use the concepts of Permutation

Given:

Total number of green bottles $=7$

Total number of blue bottles $=8$

To calculate: Number of ways in which $7$ green bottles and $8$ blue bottles can be arranged if exactly one pair of green bottle is side by side.

Consider the two green bottles as single entity.

Let's denote green bottle as $G$ and blue bottle as $B$.

Now we have to arrange total $14$ entity as two green bottles is consider as single entity.

First arrange $8$ blue bottle having equal space in between them where green bottles will filled.

There are $9$ position left in between blue bottles where green bottles can occupy.

We have total $6$ green bottles considering two green bottles as single entity.

Total number of ways of arranging $7$ green bottles and $8$ blue bottles if exactly one pair of green bottles is side by side$={}^{9}C_6=\frac{9!}{6!\left(9-6\right)!}=\frac{9!}{6!\times 3!}=\frac{9\times 8\times 7\times 6!}{6!\times 3\times 2\times 1}=84$

Hence, Number of ways in which $7$ green bottles And $8$ blue bottles can be arranged In a row if exactly one pair of green bottles is side by side is $84$ways.