Find how many numbers can be formed with the digits \[1,2,3,4,3,2,1\] so that the odd digits always occupy the odd places.
A. \[18\]
B. \[28\]
C. 6
D. \[27\]

Hint: First, find the number of odd and even numbers present in the given digits. Then, arrange the digits in even and odd places. In the end, find how many ways the digits can be arranged to get the required answer.

Formula Used: Permutation formula when repetition allowed: \[\dfrac{{n!}}{{{a_1}!{a_2}!....{a_n}!}}\]

Complete step by step solution: The given digits are \[1,2,3,4,3,2,1\].

Total number of digits: 7
Number of odd numbers: \[1,3,3,1\]: 4
Number of even numbers: \[2,4,2\]: 3

Since in the given digits 3 even numbers \[2,4,2\] are present. There are 3 even places.
And the digit 2 is repeated 2 times.
So, the number of ways of arranging the even digits at 3 places are: \[\dfrac{{3!}}{{2!}} = 3\]

Also, there are 4 even numbers \[1,3,3,1\] present and we must arrange them into 4 places.
But both numbers repeated 2 times.
So, the number of ways of arranging the odd digits at 4 places are: \[\dfrac{{4!}}{{2!2!}} = 6\]

Therefore, the number of words formed in which vowels occupy the even places are:
 \[\dfrac{{3!}}{{2!}} \times \dfrac{{4!}}{{2!2!}} = 3 \times 6\]
\[ \Rightarrow \dfrac{{3!}}{{2!}} \times \dfrac{{4!}}{{2!2!}} = 18\]

Option ‘C’ is correct

Note: Permutation shows the number of possible arrangements of the objects when the order of the arrangement of the objects matters.
If some objects are repeated, then apply the formula of the permutation for the repetition.