9 articles are to be placed in 9 boxes, one in each box, 4 of them are too big for three of the boxes. The number of possible arrangements is
$
A)\; {\text{9!}} \\
B)\; {\text{5!4!}} \\
C)\; {}^6{P_4} \times 5! \\
D)\; {\text{5!6!}} \\
$

Hint: Permutation is an ordered combination- an act of arranging the objects or numbers in the specific favourable order. The number of the permutations of the “n” objects taken “r” at the time is determined by the formula - ${}^np{}_r = \dfrac{{n!}}{{(n - r)!}}$. Use this formula in the problems where the specific favourable arrangement is
required.

Complete step by step answer:
Given that –
Total number of the articles = 9
Number of the boxes = 9
9 Articles are to be placed in 9 boxes, one in each box,
Also, given that 4 articles are very big for 3 boxes, as a result we have six boxes left out of nine boxes in which four big articles can be fit in to so that they can be chosen and can be arranged in ${}^6{P_4}$ ways. .................$(1)$
The remaining 5 articles can be arranged in 5! ways. ................$(2)$
Therefore, from the equations $\left( 1 \right)\;and\,\left( 2 \right)$
The total number of possible arrangements is $ = {}^6{P_4} \times 5!$
Therefore, the required solution - 9 articles are to be placed in 9 boxes, one in each box, 4 of them are too big for three of the boxes. The number of possible arrangements is $ = {}^6{P_4} \times 5!$.

Hence, from the given multiple options, the option C is the correct answer.

Note: In permutations, specific order and arrangement is the most important whereas, a combination is used if the certain objects are to be arranged in such a way that the order of objects is not important.