Find the number of natural number solutions t | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

$80$

Vedclass · Answer

(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 $5$.

Similar Questions

$A$ five-digit number divisible by $3$ is to be formed using the digits $0, 1, 2, 3, 4, 5$ without repetition. The total number of ways this can be done is:

Let $S$ be the set of all seven-digit numbers that can be formed using the digits $0, 1$ and $2$. For example,$2210222$ is in $S$,but $0210222$ is not in $S$. Then the number of elements $x$ in $S$ such that at least one of the digits $0$ and $1$ appears exactly twice in $x$,is equal to $....$

The remainder obtained when $1! + 2! + 95!$ is divided by $15$ is

Two packs of $52$ cards are shuffled together. The number of ways in which a man can be dealt $26$ cards so that he does not get two cards of the same suit and same denomination is

The number of ways to distribute $30$ identical candies among four children $C_{1}, C_{2}, C_{3}$ and $C_{4}$ such that $C_{2}$ receives at least $4$ and at most $7$ candies,and $C_{3}$ receives at least $2$ and at most $6$ candies,is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module