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

P_Total there are $6$ digits, from which we have to form $4$ digits number.

We need to choose $4$ digits randomly and form a number without repeating any digit.

In this process, we need to determine the number of possibilities where order does not matter.

So, we use the permutation formula and that is

${}^n{P}_r=\frac{n!}{(n-r)!},$ where n is the total number of objects and $r$ is the number of objects taken at a time.

Given, $n=6$ and $r=4$

$\therefore$ ${}^6{P}_4=\frac{6!}{(6-4)!}$

$=\frac{6!}{2!}$

$=\frac{6\times 5\times 4\times 3\times 2!}{2!}$

$=6\times 5\times 4\times 3$

$=360$

$\therefore$ ${}^6{P}_4=360$

$\therefore$ In $360$ ways the $4$-digits numbers can be form using the digits $1,2,3,7,8,9$.

From these, the even numbers must contain the last digit as $2$ or $8$.

If we fix $2$ or $8$ at $4^{th}$ place, then there are ${}^5{P}_3$ ways for each.

$\therefore$ Total number of even numbers $=2\times {}^5{P}_3$

$=2\times \frac{5!}{(5-3)!}$

$=2\times \frac{5!}{2!}$

$=2\times \frac{5\times 4\times 3\times 2\times 1}{2\times 1}$

$=120$

$\therefore$ there are total $120$ even numbers.