Garlands are formed using 4 Red roses and 4 Yellow roses of different sizes. In how many of them Red roses and Yellow roses come alternatively.
(a) 72
(b) 144
(c) 30
(d) 36

Hint: To solve this problem first we will either place the red roses or the yellow roses in the circular manner and then find the number of ways of doing so by using the formula $\left( n-1 \right)!$. After that we will count the number of places in between the placed roses and in that places we will place the other type of left roses to complete the garland and count the number of ways of doing so. After that we will finally multiply the both number of ways (as all the roses are of different sizes) to get the final answer.

Complete step by step answer:
We are given that garlands are formed by using 4 Red roses and 4 Yellow roses of different sizes and we have to find that in how many of the garlands do yellow roses and red roses come alternatively,
So to solve this problem we will use the reverse approach i.e. we will assume that the garland formed has yellow roses and red roses at alternate places and will find the number of such garlands that can be formed,
First we will place all the 4 Yellow roses in a circle to form a garland,
And this can be done in $\left( n-1 \right)!$ ways (circular permutation) where n is the number of yellow roses,
So we get,
Number of ways of placing 4 yellow roses in a circle as =,
$\begin{align}
  & =\left( n-1 \right)! \\
 & =\left( 4-1 \right)! \\
 & =\left( 3 \right)! \\
 & =3\times 2\times 1 \\
 & =6 \\
\end{align}$
Now in the formed garland there are 4 alternate places where red roses can be places,
This can be done in n! ways where n is equal to number of red roses here,
So we get,
Number of ways of placing red roses as,
= n!
= 4!
= $4\times 3\times 2\times 1$
= 24
So now total number of ways of forming such garlands we get as,
= 6 $\times $ 24
= 144

So, the correct answer is “Option B”.

Note: To solve these kinds of problems you need to know about the circular permutation that is the number of ways of arranging n things in a circle is $\left( n-1 \right)!$. So sue this some students commit mistakes by taking the first case as 4! And then they end up with the wrong answer so to solve this you need to be aware about the circular permutation.