Let n and k be positive integers such that n -k-1C2- The number of integral solutions of x1-x2- -xk-n- x1 - 1- x2 - 2- - xk - k is

Given equation is : nbsp;x_1+x_2+⋯ +x_k=n Let X_1=x_1 1, X_2=x_2 2, X_3=x_3 3, ⋯ X_k=x_k k The given equation becomes X_1+X_2+⋯ +X_k=n (1+2+3+⋯+k) X_1+X_2+⋯ + ...

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Byju's Answer

Standard XIII

Mathematics

Multinomial Theorem

Let n and k b...

Question

Let $n$ and $k$ be positive integers such that $n \geq^{k + 1} C_{2}$ . The number of integral solutions of $x_{1} + x_{2} + \dots + x_{k} = n,$ $x_{1} \geq 1, x_{2} \geq 2, \dots x_{k} \geq k$ is

(^{n -^{k} C_{2}}) C_{k}

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(^{n - 1 -^{k} C_{2}}) C_{k}

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(^{n - 1 -^{k} C_{2}}) C_{k - 1}

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(^{n + 1 -^{k} C_{2}}) C_{k - 1}

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Solution

The correct option is C $(^{n - 1 -^{k} C_{2}}) C_{k - 1}$
Given equation is : $x_{1} + x_{2} + \dots + x_{k} = n$
Let $X_{1} = x_{1} - 1, X_{2} = x_{2} - 2, X_{3} = x_{3} - 3, \dots X_{k} = x_{k} - k$
The given equation becomes $X_{1} + X_{2} + \dots + X_{k} = n - (1 + 2 + 3 + \dots + k)$
$X_{1} + X_{2} + \dots + X_{k} = n - \frac{k (k + 1)}{2}, X_{i} \geq 0$
The number of integral solutions
$=^{⎛ ⎜ ⎝ n - \frac{k (k + 1)}{2} + k - 1 ⎞ ⎟ ⎠} C_{k - 1} =^{⎛ ⎜ ⎝ n - \frac{k (k - 1)}{2} - 1 ⎞ ⎟ ⎠} C_{k - 1} [∵^{k} C_{2} = \frac{k (k - 1)}{2}] =^{(n -^{k} C_{2} - 1)} C_{k - 1}$

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Multinomial Theorem

Standard XIII Mathematics

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Let n and k be positive integers such that n≥k+1C2. The number of integral solutions of x1+x2+⋯+xk=n, x1≥1,x2≥2,⋯xk≥k is

Let $n$ and $k$ be positive integers such that $n \geq^{k + 1} C_{2}$ . The number of integral solutions of $x_{1} + x_{2} + \dots + x_{k} = n,$ $x_{1} \geq 1, x_{2} \geq 2, \dots x_{k} \geq k$ is