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 (English)

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 $x_{i} \geq a_{i}$ .

00:00 / 00:00

Conceptually Similar Problems

MathematicsProperties of Binomial CoefficientsJEE Advanced 2000Moderate

View in:English Hindi

For any positive integer $m, n$ (with $n \geq m$ ), let $(m n) =^{n} C_{m}$ . Prove that $(m n) + (m n - 1) + (m n - 2) + .... + (m m) = (m + 1 n + 1)$ . Hence or otherwise, prove that $(m n) + 2 (m n - 1) + 3 (m n - 2) + .... + (n - m + 1) (m m) = (m + 2 n + 2)$ .

MathematicsProperties of Binomial CoefficientsJEE Advanced 2003Moderate

View in:English Hindi

Prove that $2^{k} (0 n) (k n) - 2^{k - 1} (1 n) (k - 1 n - 1) + 2^{k - 2} (2 n) (k - 2 n - 2) - .... + (- 1)^{k} (k n) (0 n - k) = (k n)$

MathematicsProperties of Binomial CoefficientsJEE Advanced 1999Moderate

View in:English Hindi

Let $n$ be any positive integer. Prove that $\sum_{k = 0}^{m} \frac{( k 2 n - k )}{( n 2 n - k )} (2 n - 4 k + 1) 2^{n - 2 k} = \frac{( m n )}{( n - m 2 n - 2 m )} 2^{n - 2 m}$ for each non-negative integer $m \leq n$ .

MathematicsProperties of Binomial CoefficientsJEE Advanced 2000Easy

View in:English Hindi

For $2 \leq r \leq n$ , $(r n) + 2 (r - 1 n) + (r - 2 n) =$

(A)

(r - 1 n + 1)

(B)

2 (r + 1 n + 1)

(C)

2 (r n + 2)

(D)

(r n + 2)

Answer:D

MathematicsSolution of System of Linear Equations (Matrix Method and Cramer's Rule)JEE Main 2011Easy

View in:English Hindi

The number of values of $k$ for which the linear equations $4 x + k y + 2 z = 0$ , $k x + 4 y + z = 0$ and $2 x + 2 y + z = 0$ possess a non-zero solution is

(A)

2

(B)

1

(C)

zero

(D)

3

Answer:A

MathematicsNature of RootsJEE Advanced 1994Easy

View in:English Hindi

Let $p, q \in {1, 2, 3, 4}$ . The number of equations of the form $p x^{2} + q x + 1 = 0$ having real roots is

(A)

(B)

(C)

(D)

Answer:C

MathematicsProperties of DeterminantsJEE Main 2009Easy

View in:English Hindi

Let $a, b, c$ be such that $b (a + c) \neq = 0$ . If $a - b c a + 1 b + 1 c - 1 a - 1 b - 1 c + 1 + a + 1 a - 1 (- 1)^{n + 2} a b + 1 b - 1 (- 1)^{n + 1} b c - 1 c + 1 (- 1)^{n} c = 0$ , then the value of $n$ is:

(A)

any even integer

(B)

any odd integer

(C)

any integer

(D)

zero

Answer:B

MathematicsSolution of System of Linear Equations (Matrix Method and Cramer's Rule)JEE Main 2013Easy

View in:English Hindi

The number of values of $k$ , for which the system of equations : $(k + 1) x + 8 y = 4 k; k x + (k + 3) y = 3 k - 1$ has no solution, is

(A)

infinite

(B)

(C)

(D)

Answer:B

MathematicsTrigonometric Ratios and IdentitiesJEE Advanced 2002Easy

View in:English Hindi

The number of integral values of $k$ for which the equation $7 cos x + 5 sin x = 2 k + 1$ has a solution is

(A)

4

(B)

8

(C)

10

(D)

12

Answer:B

MathematicsCombinations and SelectionJEE Advanced 2014Easy

View in:English Hindi

Let $n_{1} < n_{2} < n_{3} < n_{4} < n_{5}$ be positive integers such that $n_{1} + n_{2} + n_{3} + n_{4} + n_{5} = 20$ . Then the number of such distinct arrangements $(n_{1}, n_{2}, n_{3}, n_{4}, n_{5})$ is

Answer:7