Question

Count the number of triples \((x, y, z)\) where \(z>\max [x, y]\) and \(1 \leq x, y, z \leq n+1\). Deduce that \(1^{2}+2^{2}+\ldots+n^{2}=C(n+1,2)+2 C(n+1,3)\)

Short answer

The number of triples is \(\sum_{k=1}^{n} k^2 = C(n+1,2) + 2C(n+1,3)\).

Step-by-step solution

Understanding the Problem

We need to count the number of ordered triples \((x, y, z)\) such that \(z > \max [x, y]\) and each of \(x, y, z\) is between \(1\) and \(n+1\), inclusive. Then, we must deduce and prove a given identity.

Rewriting Condition

The condition \(z > \max [x, y]\) implies \(z > x\) and \(z > y\). So, for a fixed \(z\), both \(x\) and \(y\) must be less than \(z\).

Counting Possibilities for Fixed z

For a given \(z\), there are \(z-1\) choices for \(x\) and \(y\) since \(x, y < z\). Therefore, there are \((z-1)^2\) choices for pairs \((x, y)\).

Summing Over All Possible z

Since \(z\) can take any value from \(2\) to \(n+1\) (as \(z=1\) cannot satisfy the condition \(z>\max[x,y]\)), we sum the squares: \[ \sum_{z=2}^{n+1} (z-1)^2 = \sum_{k=1}^{n} k^2. \]

Relation to Binomial Coefficients

The expression \(C(n+1,2)\) counts unordered pairs \((x, y)\) from \(1\) to \(n+1\). The expression \(2C(n+1,3)\) accounts for considering each such pair with a \(z\) that is greater than both. Therefore: \[ \sum_{k=1}^{n} k^2 = C(n+1,2) + 2C(n+1,3). \]

Key concepts

Ordered Triples

In mathematics, an ordered triple is simply a set of three elements presented in a specific order, denoted as \((x, y, z)\). Each position in the triple holds significance because different arrangements of the same numbers can result in entirely different triples.
For instance, \((1, 2, 3)\) is considered to be completely different from \((2, 1, 3)\) because ordering dictates the outcome when we count or consider these triples.
In the exercise given, we are concerned with the set of triples where the third element, \(z\), must be greater than the maximum of the first two elements, \(x\) and \(y\).
This requirement significantly limits our potential combinations because not every number from our range can satisfy \(z > \max[x, y]\).
For each possible \(z\), \(x\) and \(y\) must be less than \(z\), making it imperative to carefully consider each choice of \(z\) to count correctly.
By understanding the notion of ordered triples, we can better appreciate the challenge of arranging numbers under specific restrictions, which is fundamental in combinatorics.

Summation of Squares

The summation of squares refers to the mathematical expression \(1^2 + 2^2 + \, ... \, + n^2\). It describes the sum of the squares of the first \(n\) natural numbers.
This expression comes up frequently in mathematical proofs and calculations, often relating to areas under curves or discrete comparisons in algebra.
In our exercise, we use the summation of squares to count valid ordered triples by observing that for each \(z\), there are \((z-1)^2\) possible pairs \((x, y)\) since both \(x\) and \(y\) must be less than \(z\).
These comparisons lead us to the formula \(\sum_{z=2}^{n+1}(z-1)^2 = \sum_{k=1}^{n} k^2\), which simplifies our calculations by making it easier to equate it with binomial coefficients.
Understanding this summation is crucial for recognizing patterns and sum identities, which frequently appear in various fields of mathematics.

Combinatorial Proofs

Combinatorial proofs involve logical arguments that use the principle of counting to prove algebraic identities. These proofs are particularly useful because they provide a set of strategies to establish equality or equivalence between expressions through counting arguments.
In the given exercise, a combinatorial proof is utilized to prove the relationship \(1^{2}+2^{2}+ extellipsis+n^{2}=C(n+1,2)+2 C(n+1,3)\).
This identity cleverly connects the summation of squares to binomial coefficients, which are often used to count ways of choosing subsets from a larger set.

• \(C(n+1, 2)\) represents the number of ways to choose 2 elements from \(n+1\) items, essentially counting all pairs \((x, y)\).
• \(2C(n+1, 3)\) accounts for all valid triples \((x, y, z)\) where \(z\) is greater than both \(x\) and \(y\).

By seeing how these two expressions cover all possible configurations of triples, the combinatorial proof offers a holistic picture, linking algebraic manipulations directly with strategic counting methods. It's an excellent demonstration of how different mathematical domains interplay to solve complex problems.