Let A be a set of n(≥3) distinct elements. The number of triplets (x,y,z) of the elements of A in which at least two coordinates are equal is :

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Let A be a set of  n(≥3) distinct elements. The number of triplets (x,y,z) of the elements of A in which at least two coordinates are equal is :

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Given, the number of triplets (x,y,z)of the elements of A in which at least two coordinates are equal= Total number of triplets allowing repetition − Number of triplets without any repetition(i) Total number of triplets =n3 (∵Each coordinate has 3 choices and there are 3 such coordinates)(ii) Number of triplets without any repetition =n(n−1)(n−2) (∵ there and n ways of selecting first co-ordinate and (n−1) ways for selecting second co-ordinate leaving the one selected by first co-ordinate and similarly (n−2) for second)∴ The number of triplets (x,y,z) of the elements of A in which at least two coordinates are equal  =n3−n(n−1)(n−2) =3n2−2n

Question Text	Let $A$ be a set of $n (\geq 3)$ distinct elements. The number of triplets $(x, y, z)$ of the elements of $A$ in which at least two coordinates are equal is :
Answer Type	Text solution:1
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Let $A$ be a set of $n (\geq 3)$ distinct elements. The number of triplets $(x, y, z)$ of the elements of $A$ in which at least two coordinates are equal is :

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