The number of all 4 digit integers formed with exactly two distinct digits.
A. 630
B. 276
C. 576
D. 45
Answer
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Hint:We here need to find the total number of 4 digit numbers which can be formed by 2 distinct digits. For this, we will first make two different cases- one in which there is no 0 and the second in which one of the two digits is 0. We will then consider the first case and see the number of ways in which any two distinct digits can be chosen by the formula $^{n}{{C}_{r}}$ and then we will find the number of ways in which those 2 digits can be arranged at 4 distinct places and then subtract the number of cases from it in which all the 4 digits will be the same and then we will multiply them. We will do the same for the second case except we will need to choose only one-digit as 0 is already chosen and need to arrange 2 digits at three places as the first one will be the one other than 0 because if the first one becomes 0, it wouldn’t be a 4 digit number anymore. Then we will add the result of both the cases and hence we will get our answer.
Complete step by step answer:
We need to find the number of all 4 digit numbers formed with exactly two distinct digits.
Thus, if ‘a’ and ‘b’ are the two distinct digits, we need to find the numbers in which both ‘a’ and ‘b’ occurs twice, ‘a’ occurs once and ‘b’ occurs thrice, ‘b’ occurs once and ‘a’ occurs thrice but not when all the four digits are ‘a’ or ‘b’.
Now, we know that we have 10 digits to choose from-0,1,2,3,4,5,6,7,8, and 9.
Let us first only consider the numbers which don’t have any ‘0’ in them.
Thus, then we will have \[10-1=9\] out of which we need to choose any two digits.
Now, we know that number of ways of selecting ‘r’ distinct digits from ‘n’ distinct digits is given by the formula $^{n}{{C}_{r}}$ which is expanded as $\dfrac{n!}{r!\left( n-r \right)!}$ .
Here, we have 9 distinct digits out of which we need to select 2 distinct digits.
Thus, here we have:
$\begin{align}
& n=9 \\
& r=2 \\
\end{align}$
Thus, here the number of ways of selecting two digits out of 9 is given by:
$\begin{align}
& ^{n}{{C}_{r}} \\
& {{\Rightarrow }^{9}}{{C}_{2}} \\
& \Rightarrow \dfrac{9!}{2!\left( 9-2 \right)!} \\
& \Rightarrow \dfrac{9!}{2!7!} \\
\end{align}$
Now, after solving the factorials, we get:
$\begin{align}
& \dfrac{9!}{2!7!} \\
& \Rightarrow \dfrac{9\times 8\times 7!}{2!7!} \\
& \Rightarrow \dfrac{9\times 8}{2!} \\
& \Rightarrow \dfrac{9\times 8}{2\times 1} \\
& \Rightarrow 9\times 4 \\
& \Rightarrow 36 \\
\end{align}$
Thus, there are a total of 36 ways in which two distinct digits out of 1,2,3,4,5,6,7,8 and 9 can be chosen.
Now, we can further proceed to find the number of 4-digit numbers made out of these 2 digits in two ways:
1. We can either find separately the number of ways in which both the numbers will be repeated twice and then in which one will occur once and the second thrice and then in which the second one will occur once and the first one thrice.
2. We can find the total number of ways in which 4-digit numbers can be formed from these 2 digits and then subtract the number of ways in which all the 4 digits of that number are the same.
We will here use the second approach.
Now, we know that the number of ways in which ‘m’ distinct numbers can be arranged in ‘n’ different positions such that repetition is allowed is given by the formula ${{m}^{n}}$.
Here, we have 2 distinct digits that have to be arranged in 4 different positions such that repetition is allowed. Thus, we get:
$\begin{align}
& m=2 \\
& n=4 \\
\end{align}$
Hence, the number of ways of arranging two distinct digits in 4 different positions such that repetition is allowed is given as:
$\begin{align}
& {{m}^{n}} \\
& \Rightarrow {{2}^{4}} \\
& \Rightarrow 16 \\
\end{align}$
Now, let us again assume the two distinct numbers are ‘a’ and ‘b’. So, when they are arranged to form 4 digit integers, there will come a number in which all the digits will be ‘a’, i.e. ‘aaaa’ and there will come a number in which all the digits will be ‘b’, i.e. ‘bbbb’. Thus, there are a total of 2 cases in which all the digits of the 4-digit number will be the same.
Now, as mentioned above, we need to subtract these two cases from the total number of cases of the 4-digit number being formed.
Thus, the number of ways in which the two digits ‘a’ and ‘b’ can be arranged to obtain a 4-digit number is given as:
$\begin{align}
& 16-2 \\
& \Rightarrow 14 \\
\end{align}$
Now, we already established before that the number of ways of selecting any two digits from the 9 digits given as-1,2,3,4,5,6,7,8 and 9 is 36.
According to the question, we have to select any two digits and then simultaneously arrange them according to the given condition. We know that when two things happen simultaneously, the number of ways in which the simultaneous event happens is equal to the product of the number of ways in which the first event happens and the number of ways in which the second event happens.
Thus, the number of 4-digit numbers that can be formed by 2 distinct digits from 1,2,3,4,5,6,7,8 and 9 are given as:
$\begin{align}
& 36\times 14 \\
& \Rightarrow 504 \\
\end{align}$
Thus, there are a total of 504 four digit numbers which can be formed from two distinct numbers selected from 1,2,3,4,5,6,7,8 and 9.
Now, we will consider those numbers in which one of the digits is 0.
Thus, we already have one of the two digits fixed and we need to select the second one out of the remaining 9 digits.
Thus, the number of ways in which we can select one digit out of the 9 digits is given as:
$\begin{align}
& ^{9}{{C}_{1}} \\
& \Rightarrow \dfrac{9!}{1!\left( 9-1 \right)!} \\
\end{align}$
$\Rightarrow \dfrac{9!}{1!8!}$
Simplifying the factorials, we get:
$\begin{align}
& \dfrac{9!}{1!8!} \\
& \Rightarrow \dfrac{9\times 8!}{1!8!} \\
& \Rightarrow \dfrac{9}{1} \\
& \Rightarrow 9 \\
\end{align}$
Thus, there are a total of 9 ways in which 1 digit out of the 9 digits can be selected.
Now, when we start to arrange 0 and the second digit, let us assume it to be x, the first digit cannot be 0 because if it will be 0, the number will become 3 digits which is not required by us.
Thus, the first digit of the number is fixed as ‘x’.
Hence, we have to arrange the two digits, 0 and ‘x’, at 3 different positions such that they can be repeated.
Thus, the total number of ways in which the two digits, 0 and ‘x’ can be arranged at 3 different positions such that repetition is allowed is given as:
$\begin{align}
& {{2}^{3}} \\
& \Rightarrow 8 \\
\end{align}$
Now, out of these 8 numbers (which all start with x), there will be one number in which the next three digits will also be ‘x’ and hence the number will be ‘xxxx’ which according to the question is not required. Thus we need to subtract this case from a total of 8 cases.
Thus, the number of 4-digt numbers that can be formed from 0 and x is given as:
$\begin{align}
& 8-1 \\
& \Rightarrow 7 \\
\end{align}$
Now, the number of ways in which a 4-digit number can be formed such that one of them is 0 is given as:
$\begin{align}
& 9\times 7 \\
& \Rightarrow 63 \\
\end{align}$
Thus, there are a total of 63 four digit numbers which can be formed out of two distinct digits out of which one of them is 0.
Hence, the total of 4 digit numbers which can be formed from two distinct digits will be the sum of the total number of 4 digit numbers formed when neither of them is 0 and the total number of ways in which a 4 digit number that can be formed when one of them is surely 0.
Thus, we get the required number of 4 digit numbers as:
$\begin{align}
& 504+63 \\
& \Rightarrow 567 \\
\end{align}$
Hence, option (C) is the correct option.
Note:
We have here taken the second approach to find out the number of 4 digit numbers that can be formed from two distinct digits. We could have taken the first one too but we took the second one because it has less calculation and thus there will be less scope for committing a mistake. But any of the two approaches can be taken as per our convenience. They both will result in the correct answer as long as the calculations are also correct.
Complete step by step answer:
We need to find the number of all 4 digit numbers formed with exactly two distinct digits.
Thus, if ‘a’ and ‘b’ are the two distinct digits, we need to find the numbers in which both ‘a’ and ‘b’ occurs twice, ‘a’ occurs once and ‘b’ occurs thrice, ‘b’ occurs once and ‘a’ occurs thrice but not when all the four digits are ‘a’ or ‘b’.
Now, we know that we have 10 digits to choose from-0,1,2,3,4,5,6,7,8, and 9.
Let us first only consider the numbers which don’t have any ‘0’ in them.
Thus, then we will have \[10-1=9\] out of which we need to choose any two digits.
Now, we know that number of ways of selecting ‘r’ distinct digits from ‘n’ distinct digits is given by the formula $^{n}{{C}_{r}}$ which is expanded as $\dfrac{n!}{r!\left( n-r \right)!}$ .
Here, we have 9 distinct digits out of which we need to select 2 distinct digits.
Thus, here we have:
$\begin{align}
& n=9 \\
& r=2 \\
\end{align}$
Thus, here the number of ways of selecting two digits out of 9 is given by:
$\begin{align}
& ^{n}{{C}_{r}} \\
& {{\Rightarrow }^{9}}{{C}_{2}} \\
& \Rightarrow \dfrac{9!}{2!\left( 9-2 \right)!} \\
& \Rightarrow \dfrac{9!}{2!7!} \\
\end{align}$
Now, after solving the factorials, we get:
$\begin{align}
& \dfrac{9!}{2!7!} \\
& \Rightarrow \dfrac{9\times 8\times 7!}{2!7!} \\
& \Rightarrow \dfrac{9\times 8}{2!} \\
& \Rightarrow \dfrac{9\times 8}{2\times 1} \\
& \Rightarrow 9\times 4 \\
& \Rightarrow 36 \\
\end{align}$
Thus, there are a total of 36 ways in which two distinct digits out of 1,2,3,4,5,6,7,8 and 9 can be chosen.
Now, we can further proceed to find the number of 4-digit numbers made out of these 2 digits in two ways:
1. We can either find separately the number of ways in which both the numbers will be repeated twice and then in which one will occur once and the second thrice and then in which the second one will occur once and the first one thrice.
2. We can find the total number of ways in which 4-digit numbers can be formed from these 2 digits and then subtract the number of ways in which all the 4 digits of that number are the same.
We will here use the second approach.
Now, we know that the number of ways in which ‘m’ distinct numbers can be arranged in ‘n’ different positions such that repetition is allowed is given by the formula ${{m}^{n}}$.
Here, we have 2 distinct digits that have to be arranged in 4 different positions such that repetition is allowed. Thus, we get:
$\begin{align}
& m=2 \\
& n=4 \\
\end{align}$
Hence, the number of ways of arranging two distinct digits in 4 different positions such that repetition is allowed is given as:
$\begin{align}
& {{m}^{n}} \\
& \Rightarrow {{2}^{4}} \\
& \Rightarrow 16 \\
\end{align}$
Now, let us again assume the two distinct numbers are ‘a’ and ‘b’. So, when they are arranged to form 4 digit integers, there will come a number in which all the digits will be ‘a’, i.e. ‘aaaa’ and there will come a number in which all the digits will be ‘b’, i.e. ‘bbbb’. Thus, there are a total of 2 cases in which all the digits of the 4-digit number will be the same.
Now, as mentioned above, we need to subtract these two cases from the total number of cases of the 4-digit number being formed.
Thus, the number of ways in which the two digits ‘a’ and ‘b’ can be arranged to obtain a 4-digit number is given as:
$\begin{align}
& 16-2 \\
& \Rightarrow 14 \\
\end{align}$
Now, we already established before that the number of ways of selecting any two digits from the 9 digits given as-1,2,3,4,5,6,7,8 and 9 is 36.
According to the question, we have to select any two digits and then simultaneously arrange them according to the given condition. We know that when two things happen simultaneously, the number of ways in which the simultaneous event happens is equal to the product of the number of ways in which the first event happens and the number of ways in which the second event happens.
Thus, the number of 4-digit numbers that can be formed by 2 distinct digits from 1,2,3,4,5,6,7,8 and 9 are given as:
$\begin{align}
& 36\times 14 \\
& \Rightarrow 504 \\
\end{align}$
Thus, there are a total of 504 four digit numbers which can be formed from two distinct numbers selected from 1,2,3,4,5,6,7,8 and 9.
Now, we will consider those numbers in which one of the digits is 0.
Thus, we already have one of the two digits fixed and we need to select the second one out of the remaining 9 digits.
Thus, the number of ways in which we can select one digit out of the 9 digits is given as:
$\begin{align}
& ^{9}{{C}_{1}} \\
& \Rightarrow \dfrac{9!}{1!\left( 9-1 \right)!} \\
\end{align}$
$\Rightarrow \dfrac{9!}{1!8!}$
Simplifying the factorials, we get:
$\begin{align}
& \dfrac{9!}{1!8!} \\
& \Rightarrow \dfrac{9\times 8!}{1!8!} \\
& \Rightarrow \dfrac{9}{1} \\
& \Rightarrow 9 \\
\end{align}$
Thus, there are a total of 9 ways in which 1 digit out of the 9 digits can be selected.
Now, when we start to arrange 0 and the second digit, let us assume it to be x, the first digit cannot be 0 because if it will be 0, the number will become 3 digits which is not required by us.
Thus, the first digit of the number is fixed as ‘x’.
Hence, we have to arrange the two digits, 0 and ‘x’, at 3 different positions such that they can be repeated.
Thus, the total number of ways in which the two digits, 0 and ‘x’ can be arranged at 3 different positions such that repetition is allowed is given as:
$\begin{align}
& {{2}^{3}} \\
& \Rightarrow 8 \\
\end{align}$
Now, out of these 8 numbers (which all start with x), there will be one number in which the next three digits will also be ‘x’ and hence the number will be ‘xxxx’ which according to the question is not required. Thus we need to subtract this case from a total of 8 cases.
Thus, the number of 4-digt numbers that can be formed from 0 and x is given as:
$\begin{align}
& 8-1 \\
& \Rightarrow 7 \\
\end{align}$
Now, the number of ways in which a 4-digit number can be formed such that one of them is 0 is given as:
$\begin{align}
& 9\times 7 \\
& \Rightarrow 63 \\
\end{align}$
Thus, there are a total of 63 four digit numbers which can be formed out of two distinct digits out of which one of them is 0.
Hence, the total of 4 digit numbers which can be formed from two distinct digits will be the sum of the total number of 4 digit numbers formed when neither of them is 0 and the total number of ways in which a 4 digit number that can be formed when one of them is surely 0.
Thus, we get the required number of 4 digit numbers as:
$\begin{align}
& 504+63 \\
& \Rightarrow 567 \\
\end{align}$
Hence, option (C) is the correct option.
Note:
We have here taken the second approach to find out the number of 4 digit numbers that can be formed from two distinct digits. We could have taken the first one too but we took the second one because it has less calculation and thus there will be less scope for committing a mistake. But any of the two approaches can be taken as per our convenience. They both will result in the correct answer as long as the calculations are also correct.
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