Question

The total number of flags with three horizontal strips in order, which can be formed by using 2 identical red , 2 identical green and 2 identical white strips is equal to
A. 4!
B. 3 * (4!)
C. 2 * (4!)
D. None of these

Answer 1

Hint :- We had to find the total number possible combinations of flags that can be formed by using given strips and this can be easily found with the formula of total number of possible combinations = $^n{C_r}$.

Complete step-by-step answer:
As we know that there must be total three strips in every flag
So , total number of strips in a flag = 3 stripes
Now let us solve this question step by step with different possible cases.
Case 1 :- When all the horizontal strips of flag are of different colours ( i.e. red , green , , white )
As we all know that if ‘x’ number of patterns are formed with different terms than the possible number of combinations of patterns must be equal to x! .
So here also if three strips are formed with each different colour than possible number of combinations of flags = 3!
And 3! = 321 = 6 (1)
Case 2 :- When the strips of flags consist of two identical colours ( i.e. 2 identical red or 2 identical green or 2 identical white ).
Let us first of all make sure that which colour we want to fix on 2 identical strips.
Now as there are a total of 3 colours and we had to fix one of the three on identical strips this must be equal to $^3{C_1}$. (a )
Let say we had fixed red on 2 identical strips but there is one blank strip is still available
So , now let us choose the one colour that we want on the third strip.
As we had fixed red on identical strips only 2 colours are available now and we had to choose one out of these two.
So the one of the two remaining colours ( i.e. green , white ) must be $^2{C_1}$. (b)
Let's say the third strip is green.
Now the 2 identical colour ( here red ) strips can also be arranged in the horizontal rows of flag’s strips.
And for that we had to divide the total possible combinations by 2 because there are 2 fixed identical strips so this must be equal to $\dfrac{{3!}}{2}$. (c)
Let us show you all this with a short diagram

Now let us simply solve the equation (a) , (b) and (c) to get the possible combinations of flags that can be formed by two identical colour strips.
So, $^3{C_1}{ \times ^2}{C_1} \times \dfrac{{3!}}{2} = \dfrac{{3!}}{{1! \times (3 - 1)!}} \times \dfrac{{2!}}{{1! \times (2 - 1)!}} \times \dfrac{{3 \times 2 \times 1}}{2}$
And now on further solving the above equation.
$3 \times 2 \times 3 = 18$ (2)
Now adding equation (1) and (2) to get the total number of flags which can be formed
6 + 18 = 24 flags ( 24 = 4!)
Hence the correct option be A.

Note :- Whenever we come up with this type of problem we should first of all find the possible combinations when all the strips are of different colours and then we have to find the possible arrangements when the 2 strips are identical. These 2 strips can also be arranged in other ways and the remaining one strip which had different colour can also be arranged in different ways by this all we will find the combinations of flags with different strips and in the end we simply had to add the flags of different colours and the flags of identical colours and we will get the result.