A man has 3 fifty rupees notes, 4 hundred rupees notes and 6 five hundred rupees notes in his pocket. If 2 notes are taken at random, what are the odds in favour of both notes being of hundred rupee denomination?
(a) 1:12
(b) 12:1
(c) 13:1
(d) 1:13

Hint: First, before proceeding for this, we must know the value of n which is the total number of notes available irrespective of their denominations. Then, by using the concept of combination given by formula as ${}^{n}{{C}_{r}}=\dfrac{n!}{\left( n-r \right)!r!}$, we get the probability for 2 notes being 100 rupees and its opposite. Then, we know that the odds for the two 100 rupees notes to be selected from total notes is given by the ratio of above two probabilities which gives the final ratio as answer.

Complete step-by-step answer:
In this question, we are supposed to find the odds in favour of both notes being of hundred rupee denomination when man has 3 fifty rupees notes, 4 hundred rupees notes and 6 five hundred rupees notes in his pocket.
So, before proceeding for this, we must know the value of n which is the total number of notes available irrespective of their denominations.
So, we have total number of notes as:
3+4+6=13
So, we get the value of n as 13.
Then, we should know that the odds in the favour or against conditions are related to probability.
So, we need to get the probability of both the notes being selected as 100 rupee notes.
Now, by using the concept of combination given by formula as:
${}^{n}{{C}_{r}}=\dfrac{n!}{\left( n-r \right)!r!}$
Then, we know that we have total of 4 hundred rupees notes and from that we need to select only 2.
So, we get the value of n as 4 and r as 2, then we get the value of combination as:
${}^{4}{{C}_{2}}=\dfrac{4!}{\left( 4-2 \right)!2!}$
Then, by solving the above expression, we get the favourable conditions as:
$\begin{align}
  & {}^{4}{{C}_{2}}=\dfrac{4\times 3\times 2!}{2!2!} \\
 & \Rightarrow {}^{4}{{C}_{2}}=\dfrac{4\times 3}{2!} \\
 & \Rightarrow {}^{4}{{C}_{2}}=\dfrac{12}{2} \\
 & \Rightarrow {}^{4}{{C}_{2}}=6 \\
\end{align}$
So, we get the number of favourable cases as 6.
Then, by doing the same for the total number of cases as we have 13 notes and we need to select 2 notes from it.
So, we get the value of n as 13 and r as 2, then by using combination, we get:
${}^{13}{{C}_{2}}=\dfrac{13!}{\left( 13-2 \right)!2!}$
Then, by solving the above expression, we get the favourable conditions as:
$\begin{align}
  & {}^{13}{{C}_{2}}=\dfrac{13\times 12\times 11!}{11!2!} \\
 & \Rightarrow {}^{13}{{C}_{2}}=\dfrac{13\times 12}{2} \\
 & \Rightarrow {}^{13}{{C}_{2}}=13\times 6 \\
 & \Rightarrow {}^{13}{{C}_{2}}=78 \\
\end{align}$
So, we get the total number of cases as 78.
So, we get the probability of two notes being selected as 100 rupees notes from total of 13 notes as:
$\dfrac{6}{78}=\dfrac{1}{13}$
Now, we need to calculate the value of probability that two notes selected are not 100 rupees notes is given by subtracting the above probability from 1 as:
$\begin{align}
  & 1-\dfrac{1}{13}=\dfrac{13-1}{13} \\
 & \Rightarrow \dfrac{12}{13} \\
\end{align}$
Then, we know that the odds for the two 100 rupees notes be selected from total notes is given by the ratio as:
$\begin{align}
  & \dfrac{1}{13}:\dfrac{12}{13} \\
 & \Rightarrow 1:12 \\
\end{align}$
So, the odds in favour of both notes being of hundred rupees denomination is 1:12.
Hence, option (a) is correct.

Note:Now, to solve these types of the questions we need to know some of the basics of the factorial to solve the calculation in the combination formula. Now, to find the factorial of the number n, multiply the number n with (n-1) till it reaches 1. To understand let us find the factorial of 4.
$\begin{align}
  & 4!=4\times 3\times 2\times 1 \\
 & \Rightarrow 24 \\
\end{align}$