Let n and k be positive such that n ≥...

MathematicsCombinations and SelectionJEE Advanced 1996Easy

Let $n$ and $k$ be positive such that $n \geq \frac{k ( k + 1 )}{2}$ . The number of solutions $(x_{1}, x_{2}, \dots, x_{k})$ , $x_{1} \geq 1, x_{2} \geq 2, \dots, x_{k} \geq k$ , all integers, satisfying $x_{1} + x_{2} + \dots + x_{k} = n$ , is .........

Answer:

(k - 1 n - \frac{k ( k + 1 )}{2} + k - 1)

Visualized Solution (Hindi)

Understanding the Constraints

Given equation: $x_{1} + x_{2} + \dots + x_{k} = n$
Constraints: $x_{1} \geq 1, x_{2} \geq 2, \dots, x_{k} \geq k$
Total variables: $k$
Target sum: $n$

Defining New Variables .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y_{i}

Let $y_{1} = x_{1} - 1 ⟹ x_{1} = y_{1} + 1$
Let $y_{2} = x_{2} - 2 ⟹ x_{2} = y_{2} + 2$
In general, $y_{i} = x_{i} - i ⟹ x_{i} = y_{i} + i$
New constraints: $y_{1}, y_{2}, \dots, y_{k} \geq 0$

Substituting into the Equation

Substitute $x_{i} = y_{i} + i$ into the sum:
$(y_{1} + 1) + (y_{2} + 2) + \dots + (y_{k} + k) = n$
Rearrange: $(y_{1} + y_{2} + \dots + y_{k}) + (1 + 2 + \dots + k) = n$

Sum of First .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } k Integers

Sum of first $k$ natural numbers: $\sum_{i = 1}^{k} i = \frac{k ( k + 1 )}{2}$
Equation becomes: $y_{1} + y_{2} + \dots + y_{k} = n - \frac{k ( k + 1 )}{2}$
Let $N = n - \frac{k ( k + 1 )}{2}$

Applying Stars and Bars

Standard formula for $y_{1} + \dots + y_{k} = N$ with $y_{i} \geq 0$ :
Number of solutions = $(k - 1 N + k - 1)$
Substitute $N = n - \frac{k ( k + 1 )}{2}$

Final Result and Summary

Final Number of Solutions:
$(k - 1 n - \frac{k ( k + 1 )}{2} + k - 1)$
Key Takeaway: Use variable shifting $y_{i} = x_{i} - a_{i}$ to handle constraints $x_{i} \geq a_{i}$ .

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