Question

The number of ways in which $10$ persons can go in two boats so that there may be $5$ on each boat, supposing that two particular persons will not go in the same boat?
$\left( 1 \right)\dfrac{1}{2}{}^{10}{C_5}$
$\left( 2 \right)2{}^8{C_4}$
$\left( 3 \right)\dfrac{1}{2}{}^8{C_5}$
$\left( 4 \right)$ None of these

Answer 1

in order to solve this question, we will follow the formula of combination of arrangement. First, we will make one person sit in one boat in one way. Then, we will make the arrangement for remaining persons by using the formula of combination.

Complete step-by-step solution:
Since, there are a total ten persons and two boats given in the question.
So, we can make one person sit in one boat. Therefore, two persons can sit in two boats as one person in one boat and they can seat in one way.
Now, there are eight people remaining out of ten people, one person is already seated in one boat. So, we have to seat four people in each boat.
Since, the order of persons does not matter in this arrangement. So, we will use the formula of combination to make sit four persons out of eight people in one boat as:
$= {}^8{C_4}$
Since, there are two boats. Total number of ways to arrange all people in two boats is two times the number of ways to arrange four persons out of ten persons.
$= 2 \times {}^8{C_4}$
Hence, the required answer is $= 2 \times {}^8{C_4}$.

Note: Combination is a way of arrangement of some objects taken from total or whole objects without repetition. Let, there are total $n$ objects and $r$ objects needed to arrange. So, we will arrange these $r$ objects as:
$\Rightarrow {}^n{C_r}$