Let n and k be positive integers such that n - k -1 C 2-The number of integral solutions of x 1- x 2- x k - n - x 1- 1- x 2- 2- - x k - kisA- n - k C 2 C kB- n -1- k C 2 C kC- n -1- k C 2 C k -1D- n -1- k C 2 C k -1

The correct option is C (^n 1 ^kC_2)C_k 1Given 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 become ...

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Byju's Answer

Standard XII

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 is :
$=^{⎛ ⎜ ⎝ n - \frac{k (k + 1)}{2} + k - 1 ⎞ ⎟ ⎠} C_{k - 1}$
$=^{⎛ ⎜ ⎝ n - \frac{k (k - 1)}{2} - 1 ⎞ ⎟ ⎠} C_{k - 1}$ $[∵^{k} C_{2} = \frac{k (k - 1)}{2}]$
$=^{(n -^{k} C_{2} - 1)} C_{k - 1}$

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