The $5$-digit numbers must be even and positive. All the digits $1$, $2$, $3$ and $6$ must be used in each number formed. In how many ways can a $5$-digit positive even number be formed by using all of the digits $1$, $2$, $3$ and $6$?
There are only $4$ digits to be used and we have to form a $5$-digit number that means any one digit can be reused but all the digits must be included in each $5$-digit number and each number should be even.
I have tried it for $4$-digit numbers.
For the units place, we have $2$ choices, $2$ and $6$. For other $3$ places, we have $3,2,1$ choices, respectively, and, therefore, the final answer is $3 \cdot 2 \cdot 1 \cdot 2=12$ possible $4$-digit even numbers can be formed.
here is what i have tried:
for unit place we have only 2 choices,
remaining 4 places can be filled in p(4,3)i.e. 4P3=24 ways leaving one place
and that one place have 4 choices.
therefore total possible numbers can be formed is:
2*24*4=192 possible numbers
Best Answer
Any admissible string of five digits uses one of the digits $1$, $2$, $3$, $6$ exactly twice and the remaining digits exactly once. There are $4\cdot{5!\over 2!}=240$ such strings, since you can choose the digit appearing twice in four ways and then arrange the resulting quintuple in ${5!\over2!}$ ways. Exactly half of these strings have an even digit at the end. It follows that there are $120$ numbers of the required kind.