Given 4 flags of different colours, how many different signals can be generated, if a signal requires 2 flags one below the other?
$\begin{align}
  & a)4 \\
 & b)3 \\
 & c)12 \\
 & d)1 \\
\end{align}$

Hint: Now we have 4 flags of different colours. Now to find the total number of ways to create a signal we will first find the number of ways in which 4 flags can be selected among two flags. Now once we have selected two flags we will arrange those selected flags in 2! Ways.

Complete step by step answer:
Now we have 4 flags of different colours. Signals can be generated by choosing 2 flags.
Hence we will first select 2 flags out of 4 flags.
Now we know that the number of ways of selecting r objects from n objects is $^{n}{{C}_{r}}$ .
Where $^{n}{{C}_{r}}=\dfrac{n!}{(n-r)!r!}$ and $a!=a\times (a-1)\times (a-2)\times ....\times (2)\times 1$
Hence number of ways of selecting 2 flags from 4 flags is given by $^{4}{{C}_{2}}$
$^{4}{{C}_{2}}=\dfrac{4!}{(4-2)!2!}=\dfrac{4\times 3\times 2}{2\times 2}=2\times 3=6$
Hence the number of ways of choosing 2 flags out of 4 flags is 6 …………… (1)
Now we will arrange this 2 flags
We know that number of ways to arrange n objects is n!
Hence we can arrange these two flags in 2! = 2 ways ……………… (2)
Now from equation (1) and equation (2) we get that total number of ways = 6 × 2 = 12.
Hence we have the total number of signals possible is 12.

Note:
We can also think of this problem in a different manner. Let us say we have 4 flags named A, B, C and D
Now first let us say we have flag A above, then we can have B, C, D below hence we have 3 choices.
Similarly if we have a B flag above then also we have 3 choices.
Same for C and D we will have 3 choices for each.
Hence the total possible signal is 3 + 3 + 3 + 3 = 12.
Hence we have a total number of possible signals is 12.