Question

Number of ways in which $7$ green bottles and $8$ blue bottles can be arranged in a row if exactly $1$ pair of green bottles is side by side is (Assume all bottles to be a like except for the colour)
A. $84$
B. $360$
C. $504$
D. None of the above

Answer 1

Here, we have to find the ways in which $7$ green bottles and $8$ blue bottles can be arranged in a row if exactly $1$ pair of green bottles is side by side. In order to solve this question we will use the concept of permutation and combination. Permutation and combination can be defined as the methods of counting which help us to determine the number of different ways of arranging and selecting objects out of a given number of objects, without actually listing them.

Complete step by step answer:
According to the given question there are $7$ green bottles and $8$ blue bottles and we have to find the ways that they can be arranged in a row if exactly $1$ pair of green bottles is side by side. So, we will use the concept of permutation and combination. Permutation is defined as the arrangement of objects in a definite order whereas combination is defined as the selection of some or all objects from a given set of different objects when order of selection is not considered.

According to the question we have to arrange one pair of green bottles so let us consider this pair as one entity.So, there are a total $14$ units to be arranged i.e., $1$ pair of green bottles $+$ remaining $5$ green bottles $+$ $8$ blue bottles. Now we will place the $8$ blue bottles in a row, such that between each bottle there is a gap before and after it and after that we will place the six green bottles ($1$ pair of green bottles and remaining $5$ green bottles) in these nine gaps.So, there are nine gaps and six green bottles.

So, we will use the formula $^n{C_r} = \dfrac{{n!}}{{r!(n - r)!}}$. Therefore,
$\Rightarrow {}^9{C_6} = \dfrac{{9!}}{{6!\,(9 - 6)!}}$
On solving the denominator part. We get,
$\Rightarrow {}^9{C_6} = \dfrac{{9!}}{{6!\,3!}}$
on further solving. We get,
$\Rightarrow {}^9{C_6} = \dfrac{{9 \times 8 \times 7 \times 6!}}{{6!\,3!}}$
Cancelling out the equal terms in numerator and denominator. We get,
$\Rightarrow {}^9{C_6} = \dfrac{{9 \times 8 \times 7}}{{\,3 \times 2 \times 1}}$
On further simplifying. We get,
$\Rightarrow {}^9{C_6} = 3 \times 4 \times 7$
On multiplying we get,
$\Rightarrow {}^9{C_6} = 84$
Therefore, in the $84$ number of ways $7$ green bottles and $8$ blue bottles can be arranged in a row if exactly $1$ pair of green bottles is side by side.

Hence, option (A) is the correct answer.

Note: There is a difference between permutation and combination. We can define permutation as the number of ways to arrange objects and combination is defined as the number of ways to select objects. In permutation order is to be considered and in combination we do not consider order.