Question

Find the number of ways in which 3 boys and 3 girls (all are of different heights) can be arranged in a line, so that boys as well as girls among themselves are in decreasing order of height (left to right).
\[\begin{align}
  & \text{A}.\text{ 1} \\
 & \text{B}.\text{ 6}! \\
 & \text{C}.\text{ 2}0 \\
 & \text{D}.\text{ None of these} \\
\end{align}\]

Answer 1

Hint: In this question, we will use combination to get our required answer. First we will understand the logic behind this question and then solve it further. We will see that, in this question, we just have to find a pattern for 3 boys to cover 6 positions (or equivalently 3 girls to cover 6 positions) as others will take its place automatically. Formula for combination which we will use here is given as:
\[{}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}\]
Where, r items are to be chosen from n position/items.

Complete step-by-step answer:
Let us understand the meaning of combination first. Combination is a way of selecting items from a collection such that order of selection does not matter.
Now, let us analyze the question given. We are given 3 boys and 3 girls and they have to be rearranged in decreasing order of height among themselves. For solving, let's assume we have ${{B}_{1}},{{B}_{2}}\text{ and }{{B}_{3}}$ for boys in decreasing order of height which is ${{B}_{1}}$ is tallest and ${{B}_{3}}$ is shortest. Also, let's assume ${{G}_{1}},{{G}_{2}}\text{ and }{{G}_{3}}$ for girls in decreasing order of height, which is ${{G}_{1}}$ is the tallest and ${{G}_{3}}$ is shortest.
Now, we want to figure out the total pattern of boys and girls we can form. For example, BGBBGG or BBGBGG or GBBGBG are some of the patterns that can be formed. Once we have patterns, we can see that there is only one way of allotting position to the children. For example, for BGBBGG only position will be filled as ${{B}_{1}}{{G}_{1}}{{B}_{2}}{{B}_{3}}{{G}_{2}}{{G}_{3}}$. Hence, we are dependent only on finding patterns and not the ways in which they will be allotted positions. Hence, the problem reduces to finding a pattern of 3 boys within 6 positions (or equivalently 3 girls with 6 positions) as others will be filled automatically. Therefore, we have to choose 3 positions from 6 positions which will be given by ${}^{6}{{C}_{3}}$.
\[{}^{6}{{C}_{3}}=\dfrac{6!}{3!\left( 6-3 \right)!}=\dfrac{6!}{3!3!}=\dfrac{6\times 5\times 4\times 3\times 2\times 1}{3\times 2\times 1\times 3\times 2\times 1}=20\]
Hence, there are 20 ways in which children can fill 6 positions.

So, the correct answer is “Option C”.

Note: Students should take in consideration all possibilities and then conclude their final answer. Another way or we can see another approach for doing this method is:
Let’s take 6 children and count the total possible arrangements which will be 6! = 720 ways.
Now, out of these, 1 out of 3 of them will have boys in the correct order that is $\dfrac{1}{6}$.
Also, 1 out of 3 of them will have girls in the correct order that is $\dfrac{1}{6}$.
Hence, total ways becomes $720\times \dfrac{1}{6}\times \dfrac{1}{6}=20\text{ ways}$.