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(A) The total number of natural number solutions to $x_1 + x_2 = 100$ is given by the stars and bars formula $\binom{n-1}{k-1} = \binom{100-1}{2-1} =

Vedclass · Accepted Answer

$80$

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(A) The total number of natural number solutions to $x_1 + x_2 = 100$ is given by the stars and bars formula $\binom{n-1}{k-1} = \binom{100-1}{2-1} = \binom{99}{1} = 99$.Let $S$ be the set of all natural number solutions,where $|S| = 99$.Let $A$ be the set of solutions where $x_1$ is a multiple of $5$,and $B$ be the set of solutions where $x_2$ is a multiple 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 $x_1$ is a multiple of $5$,then $x_2 = 100 - x_1$ is also a multiple of $5$. Thus,$A = B$.For $x_1 \in \{5, 10, \dots, 95\}$,$x_2$ is uniquely determined as $100 - x_1$,which is also a natural number.So,$|A| = 19$ and $|B| = 19$.Since $x_1$ and $x_2$ are both multiples of $5$,the intersection $A \cap B$ occurs when $x_1$ is a multiple of $5$ and $x_2 = 100 - x_1$ is a multiple of $5$. This is true for all $19$ values in $A$.Thus,$|A \cup B| = |A| + |B| - |A \cap B| = 19 + 19 - 19 = 19$.The number of solutions where neither $x_1$ nor $x_2$ is a multiple of $5$ is $|S| - |A \cup B| = 99 - 19 = 80$.

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

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