The number of ordered triplets of positive in | vedclass

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(B) The number of positive integer solutions to the equation $x + y + z = n$ is given by the formula $\binom{n-1}{r-1}$,where $n = 100$ and $r = 3$.Us

Vedclass · Accepted Answer

$4851$

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(B) The number of positive integer solutions to the equation $x + y + z = n$ is given by the formula $\binom{n-1}{r-1}$,where $n = 100$ and $r = 3$.Using the stars and bars method for positive integers:Number of solutions $= \binom{100-1}{3-1} = \binom{99}{2}$.Calculating the value:$\binom{99}{2} = \frac{99 	imes 98}{2 	imes 1} = 99 	imes 49 = 4851$.Alternatively,using generating functions:The number of solutions is the coefficient of $x^{100}$ in $(x + x^2 + x^3 + \dots)^3$.$= 	ext{coefficient of } x^{100} 	ext{ in } x^3(1 - x)^{-3}$.$= 	ext{coefficient of } x^{97} 	ext{ in } (1 - x)^{-3}$.$= \binom{97 + 3 - 1}{3 - 1} = \binom{99}{2} = 4851$.

The number of ordered triplets of positive integers which are solutions of the equation $x + y + z = 100$ is

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