Question

Four flags of different colours are given. How many different signals can be generated if the signal requires the use of two flags, one below the other?

Answer 1

Hint: We will use the formula of combination here. . The formula is given by $C_{r}^{n}=\dfrac{n!}{r!\left( n-r \right)!}$. Where n is the total number of possibilities and r is the selected possibilities.

Complete step-by-step answer:
In the question we have four flags of different colours. Out of these we need to choose two colours and make a signal out of it. So, with the help of using the formula of combinations we will move further. The formula is given by $C_{r}^{n}=\dfrac{n!}{r!\left( n-r \right)!}$. According to the question we have a total 4 different colours and we need to make a signal by selecting any of the two out of them. Therefore, we have n = 4 and r = 2. Thus we get
$\begin{align}
  & C_{r}^{n}=\dfrac{n!}{r!\left( n-r \right)!} \\
 & \Rightarrow C_{2}^{4}=\dfrac{4!}{2!\left( 4-2 \right)!} \\
 & \Rightarrow C_{2}^{4}=\dfrac{4\times 3\times 2\times 1}{2!2!} \\
 & \Rightarrow C_{2}^{4}=\dfrac{4\times 3\times 2\times 1}{2\times 1\times 2\times 1} \\
 & \Rightarrow C_{2}^{4}=\dfrac{2\times 3\times 1}{1} \\
 & \Rightarrow C_{2}^{4}=6 \\
\end{align}$
As the two colours are also arranged by arrangement concept. So we will multiply it by 2. Therefore, we have
$\begin{align}
  & C_{2}^{4}\times 2=6\times 2 \\
 & \Rightarrow C_{2}^{4}\times 2=12 \\
\end{align}$
Hence, the total number of possibilities is done in 12 ways.

Note: Alternately we can solve it by using the concept of multiplication in which we just do multiplication until the process completes. Here we will also use the concept of addition in which we will add the possibilities of different completed works and add them together to get the total possibility. This method is done below. In this one signal is made from any two different colours out of 4 flags of different colours. And since the signal has slots one below the other. Therefore, the first slot can be filled by 4 different colours in 4 ways. Also, as we have the limitation of not repetition so we can have the second slot of the signal filled by any of the 3 flags only instead of 4. As filling up the slots is one continued work. Therefore we have a total number of possibilities given by $4\times 3=12$. Hence, the total number of possibilities is done in 12 ways.