Find the number of natural number solutions f | vedclass

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(C) The equation is $x_1 + x_2 = 100$,where $x_1, x_2 \in \mathbb{N}$.Total number of natural solutions is given by the stars and bars formula $\binom

Vedclass · Accepted Answer

$80$

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(C) The equation is $x_1 + x_2 = 100$,where $x_1, x_2 \in \mathbb{N}$.Total number of natural solutions is given by the stars and bars formula $\binom{n-1}{k-1} = \binom{100-1}{2-1} = \binom{99}{1} = 99$.We need to exclude cases where $x_1$ or $x_2$ are multiples of $5$.If $x_1$ is a multiple of $5$,then $x_1 \in \{5, 10, 15, \dots, 95\}$. There are $19$ such values.If $x_1 = 5k$,then $5k + x_2 = 100 \implies x_2 = 100 - 5k = 5(20-k)$.Since $x_2$ must be a natural number,$20-k \ge 1 \implies k \le 19$. Thus,there are $19$ solutions where $x_1$ is a multiple of $5$.Similarly,there are $19$ solutions where $x_2$ is a multiple of $5$.Note that if both $x_1$ and $x_2$ are multiples of $5$,then $x_1 = 5k$ and $x_2 = 5m$,so $5k + 5m = 100 \implies k + m = 20$.The number of natural solutions for $k+m=20$ is $\binom{20-1}{2-1} = 19$.By the Principle of Inclusion-Exclusion,the number of solutions where at least one is a multiple of $5$ is $19 + 19 - 19 = 19$.Therefore,the number of solutions where neither is a multiple of $5$ is $99 - 19 = 80$.

Find the number of natural number solutions for the equation $x_1 + x_2 = 100$,such that $x_1$ and $x_2$ are not multiples of $5$.

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