Question

Given, $9$ balls are to be placed in $9$ boxes and $5$of the balls cannot fit into $3$ small boxes. The number of ways of arranging one ball in each of the boxes is
A.$18720$
B.$18270$
C.$17280$
D.$12780$

Answer 1

The question is related to the permutation and combination concept. On considering the formula, here we choose the permutation formulas and then on substituting the known values and then by simple arithmetic operations we obtain the solution and then we choose the appropriate option.

Complete answer:
Permutation relates to the act of arranging all the members of a set into some sequence or order. In other words, if the set is already ordered, then the rearranging of its elements is called the process of permuting. Permutations occur, in more or less prominent ways, in almost every area of mathematics. They often arise when different orderings on certain finite sets are considered.
The combination is a way of selecting items from a collection, such that (unlike permutations) the order of selection does not matter. In smaller cases, it is possible to count the number of combinations
A permutation is the choice of $r$ things from a set of $n$ things without replacement and where the order matters. It is given by
${}^n{P_r} = \dfrac{{n!}}{{(n - r)!}}$
A combination is the choice of $r$ things from a set of $n$ things without replacement and where order doesn't matter. It is given by
${}^n{C_r} = \dfrac{{n!}}{{r!(n - r)!}}$
On considering the given question
Out of the $9$ different balls, $5$of the balls cannot fit into $3$ small boxes.
So those $5$balls should be arranged in the remaining $6$boxes.
$5$ Different balls can be placed in $6$boxes in ${}^6{P_5}$ ways
$\Rightarrow \dfrac{{6!}}{{1!}} = 720$ways.
Now the remaining $4$balls can be placed in $4$remaining boxes in $4!$
$\Rightarrow 24$ways.
So by using multiplication principle, the total number of ways of arranging is equal to
$\Rightarrow 720 \times 24 = 17280$
Therefore option (3) is the correct answer.

Note:
When the question is of the type arranging then in most of the cases we are using or considering the permutation formulas likewise if the question is of selection form then we are considering the combination formulas. We have to remember the factorial formula, it is given by $n! = n \times (n - 1) \times (n - 2) \times ... \times 3 \times 2 \times 1$