The number of ways \[16\] rupee and \[16\] paise coins be arranged in a line so that no two paise coins (identical) may occupy consecutive positions:
A. \[6186\]
B. \[3656\]
C. \[7575\]
D. \[6188\]

Hint: We will use both combination and permutation to solve this question. Combination is the number of selections that can be made from some or all number of things at a time. Permutation is the number of different arrangements that can be made from some or all number of things at a time.

Complete step-by-step answer:
According to the given question, we know that we are given \[16\] rupee coins and \[16\] paise coins.
We have to arrange all the coins in a line such that no two paise coins occupy consecutive positions in the line. This means that each rupee coin and each paise coin should be in alternate positions.
We first arrange the \[16\]rupee coins in one way leaving a space in between two coins.
We see that there are \[17\] spaces in between the rupee coins including the beginning and the end. We have to arrange \[16\] paise coins in between those \[17\] spaces.
Therefore, to select which of the spaces the paise coins will occupy, we will use combinations.
So, the number of ways in which the \[16\] paise coins be selected\[ = {}^{17}{C_{16}}\]
Now, we have only selected the spaces that have to be occupied but the \[16\] paise coins also get arranged in itself.
So, the number of ways in which the \[16\] paise coins can arrange in itself\[ = 16\]
Therefore, the total number of ways in which \[16\] rupee and \[16\] paise coins be arranged in a line so that no two paise coins may occupy consecutive positions
\[
   = {}^{17}{C_{16}} \times 16 \\
   = \dfrac{{17!}}{{16!\left( {17 - 16} \right)!}} \times 16 \\
   = \dfrac{{17!}}{{16!1!}} \times 16 \\
   = \dfrac{{17 \times 16!}}{{16!1!}} \times 16 \\
   = 17 \times 16 \\
   = 272 \\
\]
Thus, the answer is 272 ways.

Note: We use this method to solve this problem as the items are identical. If they had been different then the \[16\] rupee coins and the \[16\]paise coins would be arranged in \[16!\] ways each. Also, the combination part would remain the same.