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) $3\times \left( 4! \right)$
(c) $2\times \left( 4! \right)$
(d) None of these

Answer 1

Hint: First of all, consider 2 cases, first is to take all the strips of different colors and count the flags formed from them. Next is to take two strips of identical colors and one of different colors and count the flags formed by adding these values to get the required answer.

Complete step-by-step answer:
In this question, we have to find 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. We have two identical red strips, 2 identical blue strips, and 2 identical white strips and from these strips, we have to form the flag with strips placed horizontally in an order. Now, we will have two cases where all the strips of a flag would be of different color and when two strips are of identical colors. Let us see each case individually.
Case 1: When all the strips are of different colors. For example,

Red	Blue
Blue	White
White	Red

We have strips of only three colors and in this case, all should be of different colors in the flag. So, for the first strip of the flag, we have a total of 3 choices of color of strips, for the second strip of the flag, we have the remaining 2 choices and for the third strip, the choice is only 1 color remaining.
So, with 3 strips of different colors, the total number of flags formed

& =3\times 2\times 1 \\\ & =3! \\\ & =6 \\\ \end{aligned}$$ Case 2: When two strips are identical. For example White| Red ---|--- White| White Blue| Red Now, in this case, as the two strips are of the identical color. So, we will have 3 choices to select a color in which we will have two identical color strips and then out of the remaining 2 choices, we will select 1 for the third strip. So, we get the number of flags = (Ways of choosing the first color) $$\times $$ (ways of choosing the second color) $$\times $$ (Arranging number of these strips) Number of flags = 3 $$\times $$ 2 $$\times $$ (Number of the arrangement of these strips) We know that when we have (p + q) things, in which p is alike of one kind and q is alike of the second kind. So, the number of their arrangements are given by $$\dfrac{\left( p+q \right)!}{p!q!}$$ Here, we have three strips, out of which two are alike. So, we get the number of arrangements as $$\dfrac{3!}{2!}=\dfrac{3\times 2\times 1}{2\times 1}=3$$ So, we get the number of flags $$3\times 2\times \left( \dfrac{3!}{2!} \right)$$ $$3\times 2\times 3$$ $$=18$$ So, we get the total number of flags 6 + 18 = 24 flags $$=\left( 4\times 3\times 2\times 1 \right)$$ $$=4!\text{ flags}$$ So, option (a) is the right answer. Note: In this question, many students write 6! in place of 3! because they think that there are a total of 6 strips which is wrong because though we have a total of 6 strips, each of the two strips is of identical color. So, we have just 3 choices and not 6. So, 3! is correct. Also, in case 2, we have to select as well as arrange the strips separately because 2 are of identical colors. So, this should be kept in mind. Also, in this question, many students forget to add the flags of both cases. So, this must be taken care of.