Question

Find the number of elements in the relation $R:N\to N$ which is defined as $R=\left\{ \left( a,b \right)|a+b\le 10 \right\}$.

Answer 1

Hint: To solve this question, we should use the fact that the numbers a and b are natural numbers. We know the natural numbers set which is $N=\left\{ 1,2,3...... \right\}$. So, we can infer that the minimum value of a and b is 1. We should take different values of $a+b$ ranging from 1 to 10 and write the possible combinations of a and b and the sum of all the combinations gives the total number of elements in the relation R. For example, $a+b=1$ has no solution in the natural numbers because the minimum value of $a+b$ is 2 as the minimum of individual values of a and b is 1. If $a+b=2$, we get a solution which is $\left( a,b \right)=\left( 1,1 \right)$. Likewise, we should check the possible combinations for different values of a and b until $a+b=10$ and add them to get the final answer.

Complete step-by-step solution:
We are given a relation in the set of natural numbers which is $R:N\to N$ which is defined as $R=\left\{ \left( a,b \right)|a+b\le 10 \right\}$. To get the elements in the relation R, we should take different values of $a+b$ which are less than or equal to 10 and we should write different combinations possible. We know that the numbers a and b are natural numbers. So, the minimum value of a and b is 1.
Let us consider $a+b=1$.
For the given domain and range, we cannot have a solution for the equation $a+b=1$, because, for a value of $a=1$, we get $b=0$ which is not in the range.
Let us consider $a+b=2$.
The only possible solution for the above equation is when $a=1,b=1$. So, we get the number of solutions of $a+b=2$ is 1 which is $\left( 1,1 \right)$. Denoting the number with N, we can write that
$N\left( a+b=2 \right)=1$
Let us consider $a+b=3$.
The possible solutions for the above equation are when $a=2,b=1$ and $a=1,b=2$. So, we get the number of solutions of $a+b=2$ is 2 which are $\left( 2,1 \right),\left( 1,2 \right)$. We can write that
$N\left( a+b=3 \right)=2$
Let us consider $a+b=4$.
The possible solutions for the above equation are $\left( 1,3 \right),\left( 3,1 \right),\left( 2,2 \right)$ . So, we get the number of solutions of $a+b=4$ is 3. We can write that
$N\left( a+b=4 \right)=3$
Let us consider $a+b=5$.
The possible solutions for the above equation are $\left( 1,4 \right),\left( 4,1 \right),\left( 2,3 \right),\left( 3,2 \right)$ . So, we get the number of solutions of $a+b=5$ is 4. We can write that
$N\left( a+b=5 \right)=4$
Let us consider $a+b=6$.
The possible solutions for the above equation are $\left( 1,5 \right),\left( 5,1 \right),\left( 2,4 \right),\left( 4,2 \right),\left( 3,3 \right)$ . So, we get the number of solutions of $a+b=6$ is 5. We can write that
$N\left( a+b=6 \right)=5$.
If we observe the results that we got, we can get a pattern in the number of ways and the value of $a+b$. That is
$N\left( a+b=n \right)=n-1\to \left( 1 \right)$
So, we can write the total number of ways as
$N\left( a+b\le 10 \right)=N\left( a+b=2 \right)+N\left( a+b=3 \right)+N\left( a+b=4 \right)+N\left( a+b=5 \right)....N\left( a+b=10 \right)$
We can write the above value using the inference that we got in equation-1 as
$\begin{align}
  & N\left( a+b\le 10 \right)=\left( 2-1 \right)+\left( 3-1 \right)+\left( 4-1 \right)+\left( 5-1 \right)....\left( 10-1 \right) \\
 & N\left( a+b\le 10 \right)=1+2+3...9 \\
\end{align}$
We know that the sum of first n natural numbers is $1+2+3...+n=\dfrac{n\left( n+1 \right)}{2}$
Using this with the value of n = 9, we get
$N\left( a+b\le 10 \right)=1+2+3...9=\dfrac{9\left( 9+1 \right)}{2}=\dfrac{90}{2}=45$
$\therefore $ The total number of elements in the relation R is 45.

Note: An alternative way to solve this question is by using the formula that we use in permutations and combinations. The number of ways in which we can write the solution of the equation $a+b+c...r\text{ terms}=n$ when zeros are not allowed to be the values of the variables is given by ${}^{n-1}{{C}_{r-1}}$. We can write that the number of solutions for the same equation when zeros are allowed as the values of variables is ${}^{n+r-1}{{C}_{r-1}}$. The above question comes into a case where zeros are not allowed as the relation is in natural numbers. So, for every case where $a+b=n$, we get the number of solutions where $r=2$ as ${}^{n-1}{{C}_{r-1}}={}^{n-1}{{C}_{2-1}}={}^{n-1}{{C}_{1}}=\dfrac{\left( n-1 \right)!}{1!\left( n-1-1 \right)!}=n-1$. This formula also validates the inference that we had drawn by observing the results.