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Each term in the expansion of $\left ( x + y+ z \right )^{100}$ will be of the form $K*x^{m}y^{n}z^{p}$

where K is its co-efficient, also $m + n + p = 100$ ($\because$  degree of each term will be 100) and $0\leq m, n , p\leq 100$

So the number of terms in $\left ( x + y+ z \right )^{100}$ is same as number of non-negative solutions to the equation $m + n + p = 100$

$\therefore$ Number of terms $=\binom{100+3-1}{3-1}  =\binom{102}{2} = 5151$ 

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