Question

Let A be a set of $n\left( \ge 3 \right)$ distinct elements. The number of triplets (x, y, z) of the A elements in which at least two coordinates is equal to
(a) $^{n}{{P}_{3}}$
(b) ${{n}^{3}}-{{\text{ }}^{n}}{{P}_{3}}$
(c) $3{{n}^{2}}-2n$
(d) $3{{n}^{2}}\left( n-1 \right)$

Answer 1

Hint: First of all find all possible triplets that could be formed by all the elements of set A. Now subtract from it the total numbers of triplets where repetition is not allowed or no coordinates are alike to get the required answer.

Complete step-by-step answer:
In this question, we are given a set A of $n\left( \ge 3 \right)$ distinct elements. We have to find the number of triplets (x, y, z) of the A elements in which at least two coordinates are alike. Before proceeding with this question, let us know what triplets are.
As the name suggests, triplets are nothing but the set of three numbers may or may not be related to each other. For example, we have (3, 4, 5), (5, 12, 13), etc. as Pythagorean triplets because they follow Pythagoras theorem, etc. Now let us consider our question. Let us find the total number of triplets (x, y, z) that could be formed by all the n elements of set A.
In any triplet (x, y, z), x can take any one of the n values, y can also take any one of the n values, and similarly, z can also take any one of the n values. So, the total triplets (x, y, z) that can be formed by elements of set A
$=n\times n\times n$
$={{n}^{3}}.....\left( i \right)$
Now, out of the above triplets, we need to select those triplets formed by elements of set A in which at least two coordinates are alike. In other words, we need to subtract those triplets from the above total triplets in which no coordinates are alike because then only those would be left where at least two elements are alike.
So, now let us find the total triplets in which no coordinates are alike. In any triplet (x, y, z), x can take any one of the n values. Now, as the repetition is not allowed, y can only take any one of the remaining (n – 1) values. In a similar way, z can only take any one of the remaining (n – 2) values.
So, the total triplets (x, y, z) that can be formed by elements of set A without repetition
$=n\times \left( n-1 \right)\times \left( n-2 \right)$
$=n\left( n-1 \right)\left( n-2 \right)$
$=\left( {{n}^{2}}-n \right)\left( n-2 \right)$
$={{n}^{2}}\left( n-2 \right)-n\left( n-2 \right)$
$={{n}^{3}}-2{{n}^{2}}-{{n}^{2}}+2n$
$={{n}^{3}}-3{{n}^{2}}+2n.....\left( ii \right)$
We can also use the concept of permutation here, given by ${}^{n}{{P}_{r}}=\dfrac{n!}{\left( n-r \right)!}$. So, we get that the total triplets (x, y, z) that can be formed by elements of set A without repetition is
$\begin{aligned} & {}^{n}{{P}_{3}}=\dfrac{n!}{\left( n-3 \right)!} \\\ & \Rightarrow \dfrac{\left( n-3 \right)!\left( n-2 \right)\left( n-1 \right)n}{\left( n-3 \right)!} \\\ & \Rightarrow \left( n-2 \right)\left( n-1 \right)n \\\ \end{aligned}$
So, we can also write the term $n\times \left( n-1 \right)\times \left( n-2 \right)$ as ${}^{n}{{P}_{3}}\ldots \ldots \left( iii \right)$.
Now, let us find the number of triplets in which at least two coordinates are alike, so we get,
Number of triplets in which at least two coordinates are alike = (Total possible triplets) – (Total possible triplets in which no elements are alike)
By substituting the values from equation (i) and (ii), we get,
Number of triplets in which at least two coordinates are alike = ${{n}^{3}}-\left( {{n}^{3}}-3{{n}^{2}}+2n \right)=3{{n}^{2}}-2n$
By substituting the values from equation (i) and (iii), we get,
Number of triplets in which at least two coordinates are alike = ${{n}^{3}}-{}^{n}{{P}_{r}}$
Hence, option (b) and (c) are the right answers.

Note: In these types of questions, whenever we are asked to find at least a quantity, we find it by subtracting none of the quantity from the total quantity like we did in the above question. Also, in the above question, some students try to find the answer by taking two cases, that is when any 2 coordinates would be alike and when all 3 would be alike. Though this approach is also correct, it is not very easy and can often lead to confusion and wrong answers if not done properly.