How many $5$-digit positive even numbers can be formed by using all of the digits $1$, $2$, $3$ and $6$?

Question

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

Answer 1

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.

Answer 2

HINT

We can form two kinds of numbers

• $XXXX2$
• $XXXX6$

and for each one we need to consider the following sets of digits

• $\{1,2,3,6,\color{red}1\},\{1,2,3,6,\color{red}2\},\{1,2,3,6,\color{red}3\},\{1,2,3,6,\color{red}6\}$

then we can use permutations for the four X digits but we also need to take into account the pair of identical digits in each set.