For n≥ k(k+1) 2, the number of integer solutions (x₁,…,xk) with xi≥ i and x₁+·s+xk=n is ____

Answer: n- k(k+1) 2+k-1 , k-1 ,. Put yi=xi-i 0; then Σ yi=n- k(k+1) 2, and the number of non-negative solutions is n- k(k+1) 2+k-1 k-1. JEE Advanced 1996 · Quadratic Equations and Inequations — verified solution by the Vidaara Team.

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[JEE Advanced 1996] for n k k 1 2 the number of integer solutions x 1 x

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For $n\ge\frac{k(k+1)}2$, the number of integer solutions $(x_1,\dots,x_k)$ with $x_i\ge i$ and $x_1+\cdots+x_k=n$ is ____

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VAVidaara Admin ✓ Vidaara Team ✓ Accepted · 2d ago ▲ 0

Answer: $\dbinom{\,n-\frac{k(k+1)}2+k-1\,}{\,k-1\,}$.

Put $y_i=x_i-i\ge0$; then $\sum y_i=n-\frac{k(k+1)}2$, and the number of non-negative solutions is $\binom{n-\frac{k(k+1)}2+k-1}{k-1}$.

JEE Advanced 1996 · Quadratic Equations and Inequations — verified solution by the Vidaara Team.

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