Question

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

Answer 1

Hint: According to given in the question we have to determine the total number of flags with three horizontal strips in order, which can be formed using 2 identical red, 2 identical green, and 2 identical white strips. So, first of all as mentioned in the question that there are three strips of different colours like red, green and white in colours so, we have to determine the number of ways to arranging all of the three colours which can be done with the help of the formula which is as mentioned below:

Formula used:
$ \Rightarrow n! = n(n - 1)(n - 2)........................(A)$
Hence, with the help of the formula (A) above, we can determine the total number of ways to arrange all the three colour strips.
Now, we have to determine the number of ways when the two colours strips are of same colours which can be red, green or white with the help of the formula as mentioned below:
$ \Rightarrow c_r^n = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}.................(B)$
Where, n is the total number of choices and r is the choices we required.
Now, we have to determine if one colour strip is chosen at random from the three different colour strips which can be determined with the help of the formula (B) as mentioned above.
Now, to obtain the required number of ways we have to add the both of the ways for arranging the three colours strips and the number of ways to choose the strip of any one colour of the strip.

Complete step-by-step answer:
Step 1: First of all as mentioned in the question that there are three strips of different colours like red, green and white in colours so, we have to determine the number of ways to arranging all of the three colours which can be done with the help of the formula (A) which is as mentioned in the solution hint. Hence,
$
   \Rightarrow 3! = 3 \times 2 \times 1 \\
   \Rightarrow 3! = 6 \\
 $
Step 2: Now, we have to determine the number of ways when the two colours strips are of the same colours which can be red, green or white with the help of the formula (B) as mentioned in the solution hint. Hence,
$
   = c_1^2 \times \dfrac{{3!}}{2} \\
   = \dfrac{{2!}}{{1!\left( {2 - 1} \right)!}} \times \dfrac{{3 \times 2 \times 1}}{2} \\
   = \dfrac{{2 \times 1}}{{1!\left( 1 \right)!}} \times 3 \\
   = 2 \times 3 \\
   = 6 \\
 $
Step 3: Now, we have to determine if one colour strip is chosen at random from the three different colour strips which can be determined with the help of the formula (B) as mentioned in the solution hint. Hence,
\[
   = c_1^3 \\
   = \dfrac{{3!}}{{1!\left( {3 - 1} \right)!}} \\
   = \dfrac{{3 \times 2!}}{{1! \times 2!}} \\
   = 3 \\
 \]
Step 4: Now, we have to multiply the both of the ways as obtained in the solution step 2 and 3 hence,
\[
   = 6 \times 3 \\
   = 18 \\
 \]
Step 5: Now, to obtain the required number of ways we have to add the both of the ways for arranging of the three colours strips and the number of ways to choose the strip of any one colour of the strip as mentioned in the solution hint. Hence,
$
   = 6 + 18 \\
   = 24 \\
 $
Step 6: Now, we can also represent the required number 24 in the form of factorial with the help of the formula (A) as mentioned in the solution hint. Hence,
$ \Rightarrow 4 \times 3 \times 2 \times 1 = 4!$

Hence, with the help of the formulas (A) and (B) which are as mentioned in the solution hint we have determined the total number of ways which is $4!$. Therefore option (A) is correct.

Note:
To convert the obtained number in the form of factorial we have to use the formula $n! = n(n - 1)(n - 2)$ where n is the number which we want to convert in the form of factorial. It is necessary that we have to divide the obtained number of ways to arrange all the three colours flag which is 3! By 2 because two same colours can be chosen while forming the flag.