If ${{n}_{1}}\ and\ {{n}_{2}}$ are five – digit numbers, find the total number of ways of forming ${{n}_{1}}\ and\ {{n}_{2}}$. So that these numbers can be added without carrying at any stage.
Answer
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Hint: We will first start by seeing a fact that if two digits on adding is greater than 9 then there will be no carry on in the number. Then we will use this fact to find the cases for each of the digits and then using multiplication principle will multiply each case to find the total cases.
Complete step-by-step answer:
Now, we will first let the number \[{{n}_{1}}={{x}_{1}}{{x}_{2}}{{x}_{3}}{{x}_{4}}{{x}_{5}}\] where \[{{x}_{i}}\] for i = 1 to 5 is digit of the number ${{n}_{1}}$. We have the number ${{n}_{2}}$ as \[{{n}_{2}}={{y}_{1}}{{y}_{2}}{{y}_{3}}{{y}_{4}}{{y}_{5}}\] where \[{{y}_{i}}\] for i = 1 to 5 is digit of the number ${{n}_{2}}$.
Now, we have been given that by adding ${{n}_{1}}\ and\ {{n}_{2}}$ there should be no carry on. So, the sum of each digit at each place must be less than or equal to 9 this means,
\[{{n}_{1}}={{x}_{1}}{{x}_{2}}{{x}_{3}}{{x}_{4}}{{x}_{5}}\]
\[{{n}_{2}}={{y}_{1}}{{y}_{2}}{{y}_{3}}{{y}_{4}}{{y}_{5}}\]
\[\Rightarrow {{x}_{5}}+{{y}_{5}}\le 9\]
Now, if we take \[{{x}_{5}}=0\] then \[{{y}_{5}}\le 9\]. So, there are 10 possible ${{y}_{5}}$ for \[{{x}_{5}}=0\].
Similarly, for \[{{x}_{5}}=1\] we have \[{{y}_{5}}\le 8\]. Therefore, there are 9 possible ${{y}_{5}}$ for \[{{x}_{5}}=1\].
Similarly, we can see that possible number of values of ${{y}_{5}}$ for ${{x}_{5}}=2,3..........9\ is\ 8,7............1$ respectively.
Therefore, the total possible cases for \[{{x}_{5}}+{{y}_{5}}\le 9\] is \[1+2+3+.........+10\].
Now, we know that the sum of the first n natural number is $\dfrac{n\left( n+1 \right)}{2}$.
\[\Rightarrow \] Total cases for \[{{x}_{5}}+{{y}_{5}}\le 9=\dfrac{10+\left( 10+1 \right)}{2}\]
\[=55\]
Now, we have the same number of cases for \[{{x}_{4}}+{{y}_{4}}\le 9,{{x}_{3}}+{{y}_{3}}\le 9,{{x}_{2}}+{{y}_{2}}\le 9\].
But for \[{{x}_{1}}+{{y}_{1}}\le 9\] we have to notice that ${{x}_{1}}={{y}_{1}}\ne 0$ because then the number of questions will be less than 5. So, we have for \[{{x}_{1}}=1,{{y}_{1}}\le 8\]. So, there are 8 cases and similarly for \[{{x}_{1}}=2,3.......9\] the cases are \[7,6..........1\].
So, the total cases for \[{{x}_{1}}+{{y}_{1}}\le 9\] is \[8+7+6+5+......+1\].
So, we have total cases for \[{{x}_{1}}+{{y}_{1}}\le 9\] as $\dfrac{8\left( 9 \right)}{2}=36$.
Now, we have to find the total ways in which ${{n}_{1}}\ and\ {{n}_{2}}$ can be formed. So, multiplying the cases for each x, y we have,
\[\begin{align}
& 36\times 55\times 55\times 55\times 55 \\
& \Rightarrow 36\times {{\left( 55 \right)}^{4}} \\
\end{align}\]
Hence, the total possible ways are \[36\times {{\left( 55 \right)}^{4}}\].
Note: It is important to note that we have to find the cases for \[{{x}_{1}}+{{y}_{1}}\le 9\] separately because the first digit of ${{n}_{1}}\ and\ {{n}_{2}}$ can’t be zero. Therefore, ${{x}_{1}}\ and\ {{y}_{1}}$ can’t be zero. Hence, we have to calculate it separately.
Complete step-by-step answer:
Now, we will first let the number \[{{n}_{1}}={{x}_{1}}{{x}_{2}}{{x}_{3}}{{x}_{4}}{{x}_{5}}\] where \[{{x}_{i}}\] for i = 1 to 5 is digit of the number ${{n}_{1}}$. We have the number ${{n}_{2}}$ as \[{{n}_{2}}={{y}_{1}}{{y}_{2}}{{y}_{3}}{{y}_{4}}{{y}_{5}}\] where \[{{y}_{i}}\] for i = 1 to 5 is digit of the number ${{n}_{2}}$.
Now, we have been given that by adding ${{n}_{1}}\ and\ {{n}_{2}}$ there should be no carry on. So, the sum of each digit at each place must be less than or equal to 9 this means,
\[{{n}_{1}}={{x}_{1}}{{x}_{2}}{{x}_{3}}{{x}_{4}}{{x}_{5}}\]
\[{{n}_{2}}={{y}_{1}}{{y}_{2}}{{y}_{3}}{{y}_{4}}{{y}_{5}}\]
\[\Rightarrow {{x}_{5}}+{{y}_{5}}\le 9\]
Now, if we take \[{{x}_{5}}=0\] then \[{{y}_{5}}\le 9\]. So, there are 10 possible ${{y}_{5}}$ for \[{{x}_{5}}=0\].
Similarly, for \[{{x}_{5}}=1\] we have \[{{y}_{5}}\le 8\]. Therefore, there are 9 possible ${{y}_{5}}$ for \[{{x}_{5}}=1\].
Similarly, we can see that possible number of values of ${{y}_{5}}$ for ${{x}_{5}}=2,3..........9\ is\ 8,7............1$ respectively.
Therefore, the total possible cases for \[{{x}_{5}}+{{y}_{5}}\le 9\] is \[1+2+3+.........+10\].
Now, we know that the sum of the first n natural number is $\dfrac{n\left( n+1 \right)}{2}$.
\[\Rightarrow \] Total cases for \[{{x}_{5}}+{{y}_{5}}\le 9=\dfrac{10+\left( 10+1 \right)}{2}\]
\[=55\]
Now, we have the same number of cases for \[{{x}_{4}}+{{y}_{4}}\le 9,{{x}_{3}}+{{y}_{3}}\le 9,{{x}_{2}}+{{y}_{2}}\le 9\].
But for \[{{x}_{1}}+{{y}_{1}}\le 9\] we have to notice that ${{x}_{1}}={{y}_{1}}\ne 0$ because then the number of questions will be less than 5. So, we have for \[{{x}_{1}}=1,{{y}_{1}}\le 8\]. So, there are 8 cases and similarly for \[{{x}_{1}}=2,3.......9\] the cases are \[7,6..........1\].
So, the total cases for \[{{x}_{1}}+{{y}_{1}}\le 9\] is \[8+7+6+5+......+1\].
So, we have total cases for \[{{x}_{1}}+{{y}_{1}}\le 9\] as $\dfrac{8\left( 9 \right)}{2}=36$.
Now, we have to find the total ways in which ${{n}_{1}}\ and\ {{n}_{2}}$ can be formed. So, multiplying the cases for each x, y we have,
\[\begin{align}
& 36\times 55\times 55\times 55\times 55 \\
& \Rightarrow 36\times {{\left( 55 \right)}^{4}} \\
\end{align}\]
Hence, the total possible ways are \[36\times {{\left( 55 \right)}^{4}}\].
Note: It is important to note that we have to find the cases for \[{{x}_{1}}+{{y}_{1}}\le 9\] separately because the first digit of ${{n}_{1}}\ and\ {{n}_{2}}$ can’t be zero. Therefore, ${{x}_{1}}\ and\ {{y}_{1}}$ can’t be zero. Hence, we have to calculate it separately.
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