In how many ways can \[12\] balls be divided between \[2\] boys, one receiving \[5\] and the other \[7\] balls?
1.\[1080\]
2.\[1184\]
3.\[1584\]
4. none of these

Hint: In order to find the number of ways in which \[12\] balls can be divided between \[2\] boys, one receiving \[5\] and the other \[7\] balls, firstly we will be considering the total number of balls and then the ways in which the balls can be given. Calculating this will give us the required answer.

Complete step by step answer:
Now let us briefly discuss the combinations. Combinations are selection of items from a group of items when the order of the selection is not considered. Combination simply deals with the selection. The notation of the combination is \[^{n}{{C}_{r}}\]. The formula for finding the number of combinations is \[^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}\]. There are two basic principles of combinations. They are: Fundamental Principle of Counting and Addition Principle.
Now let us start solving the given problem.
We are given,
The total number of balls is \[12\] and the number of boys is \[2\].
Now let calculate the number of ways in which the given condition can be satisfied. We get,
\[\Rightarrow \dfrac{12!}{\left( 5!\times 7! \right)}\times 2!\]
Upon further solving this, we obtain
\[\begin{align}
  & \Rightarrow \dfrac{12!}{\left( 5!\times 7! \right)}\times 2! \\
 & \Rightarrow 11\times 9\times 8\times 2=1584 \\
\end{align}\]

So, the correct answer is “Option 3”.

Note: While solving for the number of ways, we must not forget to consider all the possible ways. We have multiplied with \[2!\] in the above problem because we don’t know which boy would be receiving \[5\] and the other \[7\] balls, that could be vice versa too. Not considering all the ways would be the commonly committed error.