Question

In how many ways 4-digits numbers can be formed using the digits 1, 2, 3, 7, 8, 9 without repetition? How many of these are even numbers?

Answer 1

Hint – In this particular question use the concept that the number of ways to select r distinct object out of n distinct object is ${}^n{C_r}$ and the number of arrangements of n different objects is n! so use these concepts to reach the solution of the question.

Complete step by step solution:
Given digits:
1, 2, 3, 7, 8, 9.
As we see all are different digits and there are a total 6 digits available.
Now we have to make a four digit number without repetition.
So as we know that the number of ways to select r distinct object out of n distinct object is ${}^n{C_r}$.
So the number of ways to select 4 digits out of 6 different digits is ${}^6{C_4}$.
Now we have to arrange these digits.
As we know that the number of arrangements of n different objects = n!
So the number of arrangements of 4 digits = 4!
So the total number of 4 digit numbers is the product of the above two calculated values.
So the total number of 4 digit number without repetition = ${}^6{C_4} \times 4!$
Now as we know that ${}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$ so use this formula we have,
So the total number of 4 digit number without repetition = $\dfrac{{6!}}{{4!\left( {6 - 4} \right)!}} \times 4!$
$\Rightarrow \dfrac{{6.5.4!}}{{4!.2.1}} \times 4.3.2.1$
$\Rightarrow 15 \times 24 = 360$
So the total number of 4 digit numbers from the given digits are 360.
Now we have to find out how many of these are even.
As we all know that for an even number the last digit of the number should be even.
So in the given digits there are only 2 even digits which are 2 and 8.
Case – 1: When last digit is 2.
So we have to choose another 3 digit from the remaining (6 – 1) = 5 digits to complete the 4 digit number.
So the number of ways to select 3 digits out of 5 different digits is ${}^5{C_3}$.
Now we have to arrange these digits.
As we know that the number of arrangements of n different objects = n!
So the number of arrangements of 3 digits = 3!
So the number of even 4 digit numbers when the last digit is fixed i.e. 2 is the product of the above two calculated values.
So the number of even 4 digit number when the last digit is fixed i.e. 2 = ${}^5{C_3} \times 3!$
Now as we know that ${}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$ so use this formula we have,
So the number of even 4 digit number when the last digit is fixed i.e. 2, = $\dfrac{{5!}}{{3!\left( {5 - 3} \right)!}} \times 3!$
$\Rightarrow \dfrac{{5.4.3!}}{{3!.2.1}} \times 3.2.1$
$\Rightarrow 10 \times 6 = 60$
So when last digit is 2 the number of even 4 digit numbers are 60
Similarly when the last digit is 8 the number of even 4 digit numbers are 60.
So the total number of even 4 digit numbers from the given digit = 60 + 60 = 120.
So this is the required answer.

Note – Whenever we face such types of questions the key concept we have to remember is that always recall the formula of combinations which is stated above, we can also solve this question by using permutations in permutations both arrangements and combinations are taken at a time i.e. the number of 4 digit numbers out of 6 numbers is ${}^6{P_4} = \dfrac{{6!}}{{\left( {6 - 4} \right)!}} = \dfrac{{6!}}{{2!}} = \dfrac{{720}}{2} = 360$.