Question

eg: let n be a positive int. p.t no of ways of expressing $n = a+b$ where a & b are int with $a\geq b\geq 0$ = expressing $n = c+2d$ c&d are non -ve int

Explanation:

Step1: Define first set of pairs

Let \(S_1 = \{(a,b) \mid n = a+b, a \geq b \geq 0, a,b \in \mathbb{Z}\}\)

Step2: Define second set of pairs

Let \(S_2 = \{(c,d) \mid n = c+2d, c,d \geq 0, c,d \in \mathbb{Z}\}\)

Step3: Map \(S_1\) to \(S_2\)

Case 1: If \(a = b\), then \(n=2a\). Set \(c=0, d=a\).
Case 2: If \(a > b\), write \(a = b + k, k \geq 1\). Then \(n=2b+k\). Set \(c=k, d=b\).

Step4: Map \(S_2\) to \(S_1\)

Case 1: If \(c=0\), \(n=2d\). Set \(a=d, b=d\).
Case 2: If \(c>0\), set \(a = d + c, b = d\), so \(a+b=2d+c=n\) and \(a>b\).

Step5: Verify bijection

Both maps are inverses, so \(|S_1|=|S_2|\).

Answer:

The number of non-negative integer pairs \((a,b)\) with \(a \geq b\) such that \(n=a+b\) is equal to the number of non-negative integer pairs \((c,d)\) such that \(n=c+2d\).