Question

Number of four digit numbers that are divisible by \[2\] if repetition of digits is not allowed is?

Answer 1

Hint: Here we use the concept of combination without repetition in which the total number of choices decreases by one in every step. We divide the numbers into odd and even and fix one even for the last digit so the number can be divisible by \[2\].
* Number of ways to choose \[r\] objects from total of \[n\] objects is given by \[^n{C_r} = \dfrac{{n!}}{{(n - r)!r!}}\]

Complete step-by-step answer:
We know any number is divisible by \[2\] if it has \[0,2,4,6,8\] in its place.
\[ \Rightarrow \] The numbers ending with the digits \[0,2,4,6,8\] are even and so are divisible by \[2\].
So we take all the possibilities of choosing digits by keeping in mind that last digit should be from \[0,2,4,6,8\].
We have five odd numbers and five even numbers from \[0\] to \[10\].
Odd numbers are \[1,3,5,7,9\] and even numbers are \[0,2,4,6,8\]
We have four possibilities to choose a number.
First: First three digits are all odd.
So we can choose the first three digits from a total of five odd digits and the last digit can be chosen from five even digits. Whereas choosing the first three digits repetition is not allowed so one digit is chosen from five digits, then one digit is chosen from four digits and one digit is chosen from three digits.
Then number of ways to choose the number is \[^5{C_1}{ \times ^4}{C_1}{ \times ^3}{C_1}{ \times ^5}{C_1}\]
From the formula \[^n{C_r} = \dfrac{{n!}}{{(n - r)!r!}}\],
When we put \[r = 1\] we get \[^n{C_1} = \dfrac{{n!}}{{(n - 1)!1!}} = \dfrac{{n(n - 1)!}}{{(n - 1)!}} = n\]
\[{ \Rightarrow ^5}{C_1} = 5{,^4}{C_1} = 4{,^3}{C_1} = 3\]
Substituting the values
\[{ \Rightarrow ^5}{C_1}{ \times ^4}{C_1}{ \times ^3}{C_1}{ \times ^5}{C_1} = 5 \times 4 \times 3 \times 5 = 300\] … (i)
Second: One of the first three digits is even.
While choosing the first three digits repetition is not allowed so one digit is chosen from five even digits, then one digit is chosen from five odd digits and one digit is chosen from four odd digits and last digit is chosen from four even digits.
Then number of ways to choose the number is \[^5{C_1}{ \times ^4}{C_1}{ \times ^5}{C_1}{ \times ^4}{C_1}\]
From the formula \[^n{C_r} = \dfrac{{n!}}{{(n - r)!r!}}\],
When we put \[r = 1\] we get \[^n{C_1} = \dfrac{{n!}}{{(n - 1)!1!}} = \dfrac{{n(n - 1)!}}{{(n - 1)!}} = n\]
\[{ \Rightarrow ^5}{C_1} = 5{,^4}{C_1} = 4\]
Substituting the values
\[{ \Rightarrow ^5}{C_1}{ \times ^4}{C_1}{ \times ^5}{C_1}{ \times ^4}{C_1} = 5 \times 4 \times 5 \times 4 = 400\] … (ii)
Third: Two of the first three digits are even.
While choosing the first three digits repetition is not allowed so one digit is chosen from five even digits, one digit is chosen from four even digits, one is chosen from five odd digits and last digit is chosen from three even digits.
Then number of ways to choose the number is \[^5{C_1}{ \times ^4}{C_1}{ \times ^5}{C_1}{ \times ^3}{C_1}\]
From the formula \[^n{C_r} = \dfrac{{n!}}{{(n - r)!r!}}\],
When we put \[r = 1\] we get \[^n{C_1} = \dfrac{{n!}}{{(n - 1)!1!}} = \dfrac{{n(n - 1)!}}{{(n - 1)!}} = n\]
\[{ \Rightarrow ^4}{C_1} = 4{,^5}{C_1} = 5{,^3}{C_1} = 3\]
Substituting the values
\[^5{C_1}{ \times ^4}{C_1}{ \times ^5}{C_1}{ \times ^3}{C_1} = 5 \times 4 \times 5 \times 3 = 300\] … (iii)
Fourth: all of the first three digits are even.
While choosing the first three digits repetition is not allowed so one digit is chosen from five even digits, then one digit is chosen from four even digits, one is chosen from three even digits and last is chosen from two even digits.
Then number of ways to choose the number is \[^5{C_1}{ \times ^4}{C_1}{ \times ^3}{C_1}{ \times ^2}{C_1}\]
From the formula \[^n{C_r} = \dfrac{{n!}}{{(n - r)!r!}}\],
When we put \[r = 1\] we get \[^n{C_1} = \dfrac{{n!}}{{(n - 1)!1!}} = \dfrac{{n(n - 1)!}}{{(n - 1)!}} = n\]
\[{ \Rightarrow ^5}{C_1} = 5{,^4}{C_1} = 4{,^3}{C_1} = 3{,^2}{C_1} = 2\]
Substituting the values
\[{ \Rightarrow ^5}{C_1}{ \times ^4}{C_1}{ \times ^3}{C_1}{ \times ^2}{C_1} = 5 \times 4 \times 3 \times 2 = 120\] … (iv)
Add equations (i), (ii), (iii) and (iv)
Now we add all the possible ways \[ = 300 + 400 + 300 + 120 = 1120\]
So, the number of four digit numbers that are divisible by \[2\] is \[1120\].
Since, we are choosing the first three digits, so we can interchange these three digits. So, the number of ways to write three digits in three places is given by \[3! = 3 \times 2 \times 1 = 6\]
Therefore, the number of four digit numbers that are divisible by \[2\] is \[1120 \times 6 = 6720\].

Note: Students many times make the mistake of directly applying the combination formula and don’t use the concept of repetition which is wrong, keep in mind to avoid repetition of digits and reduce one from the value of \[n\].