Question

Number of ways 6 rings can be worn on four fingers of one hand?
\[\begin{align}
  & \text{A}.\text{ 4}0\text{95} \\
 & \text{B}.\text{ 4}0\text{96} \\
 & \text{C}.\text{ 4}0\text{97} \\
 & \text{D}.\text{ 4}0\text{98} \\
\end{align}\]

Answer 1

Hint: In this question, we are given 6 rings and four fingers and we have to find the number of ways 6 rings can be worn in four fingers of one hand. As we can see, this is the sum of permutation. So, we will first understand the logic behind sum and then solve it to found our required solution.

Complete step-by-step answer:
Before solving our question, let us first understand the meaning of permutation. Permutation relates to the act of arranging all the members of a set into some sequence or order.
In this case, we have to arrange six rings in the four fingers. So, let's understand the logic behind this. Let us suppose, we have six rings as ${{R}_{1}},{{R}_{2}},{{R}_{3}},{{R}_{4}},{{R}_{5}}\text{ and }{{\text{R}}_{\text{6}}}$ and four fingers as ${{F}_{1}},{{F}_{2}},{{F}_{3}}\text{ and }{{\text{F}}_{\text{4}}}$. As we can see for ring ${{R}_{1}}$ it can be put on any finger ${{F}_{1}},{{F}_{2}},{{F}_{3}}\text{ or }{{\text{F}}_{\text{4}}}$. Hence, we have four ways for ${{R}_{1}}$.
\[\Rightarrow {{R}_{1}}\begin{matrix}
   \nearrow \\
   \to \\
   \to \\
   \searrow \\
\end{matrix}\begin{matrix}
   {{F}_{1}} \\
   {{F}_{2}} \\
   {{F}_{3}} \\
   {{F}_{4}} \\
\end{matrix}\]
For ${{R}_{2}}$ it can also be put on any finger ${{F}_{1}},{{F}_{2}},{{F}_{3}}\text{ or }{{\text{F}}_{\text{4}}}$. Hence, we have four ways for ${{R}_{2}}$.
\[\Rightarrow {{R}_{2}}\begin{matrix}
   \nearrow \\
   \to \\
   \to \\
   \searrow \\
\end{matrix}\begin{matrix}
   {{F}_{1}} \\
   {{F}_{2}} \\
   {{F}_{3}} \\
   {{F}_{4}} \\
\end{matrix}\]
Similarly, for ${{R}_{3}},{{R}_{4}},{{R}_{5}}\text{ and }{{\text{R}}_{\text{6}}}$. There are four ways for each ${{R}_{3}},{{R}_{4}},{{R}_{5}}\text{ and }{{\text{R}}_{\text{6}}}$.
\[\begin{align}
  & \Rightarrow {{R}_{3}}\begin{matrix}
   \nearrow \\
   \to \\
   \to \\
   \searrow \\
\end{matrix}\begin{matrix}
   {{F}_{1}} \\
   {{F}_{2}} \\
   {{F}_{3}} \\
   {{F}_{4}} \\
\end{matrix} \\
 & \Rightarrow {{R}_{4}}\begin{matrix}
   \nearrow \\
   \to \\
   \to \\
   \searrow \\
\end{matrix}\begin{matrix}
   {{F}_{1}} \\
   {{F}_{2}} \\
   {{F}_{3}} \\
   {{F}_{4}} \\
\end{matrix} \\
 & \Rightarrow {{R}_{5}}\begin{matrix}
   \nearrow \\
   \to \\
   \to \\
   \searrow \\
\end{matrix}\begin{matrix}
   {{F}_{1}} \\
   {{F}_{2}} \\
   {{F}_{3}} \\
   {{F}_{4}} \\
\end{matrix} \\
 & \Rightarrow {{R}_{6}}\begin{matrix}
   \nearrow \\
   \to \\
   \to \\
   \searrow \\
\end{matrix}\begin{matrix}
   {{F}_{1}} \\
   {{F}_{2}} \\
   {{F}_{3}} \\
   {{F}_{4}} \\
\end{matrix} \\
\end{align}\]
Hence, for every finger we have four ways each.
Therefore, for six fingers, ways becomes \[\Rightarrow 4\times 4\times 4\times 4\times 4\times 4=4096\]
Hence, there are 4096 ways in which six fingers can be worn on four fingers of one hand.

So, the correct answer is “Option B”.

Note: Students should always try to draw small diagrams for understanding concepts clearly. They should not get confused with whether to add or multiply for all possible ways. Addition is used when we use 'or' but multiplication is used when we use 'and'. Also, there can arise confusion while knowing the difference between permutation and combination. Permutation tells us how to arrange items and combination tells us how to select items.