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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
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  • $\begingroup$ Please edit the question to show us what you have tried. Can you solve an easier similar problem? $\endgroup$
    – Ethan Bolker
    Jun 29, 2018 at 16:10
  • $\begingroup$ @ChristianBlatter that means non negative; i introduced +ve number just to be safe from some people who will form negative number too $\endgroup$
    – Bibek Ghimire
    Jun 30, 2018 at 1:08
  • $\begingroup$ Please read this tutorial on how to typeset mathematics on this site using MathJax. $\endgroup$
    – N. F. Taussig
    Jun 30, 2018 at 9:02
  • $\begingroup$ It matters whether or not the units digit is the repeated digit. $\endgroup$
    – N. F. Taussig
    Jul 5, 2018 at 8:44
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    $\begingroup$ As you observed, the units digit can be selected in two ways. If the units digit is the repeated digit, then the first four digits are all different, so they can be arranged in $4!$ ways, giving $2 \cdot 4! = 48$ such arrangements. If the units digit is not the repeated digit, the repeated digit can be selected in three ways, its positions can be selected in $\binom{4}{2}$ ways (since it cannot be placed in the units digit), and the remaining two positions can be filled in $2!$ ways, giving $2 \cdot \binom{3}{1}\binom{4}{2} \cdot 2! = 72$ such arrangements. $\endgroup$
    – N. F. Taussig
    Jul 6, 2018 at 9:00

2 Answers 2

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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.

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  • $\begingroup$ i have also tried to find the solution but the answer is different from yours. I want you to see my answer and correct my misconception about this. Where should i submit my answer so that you can see? in the comment box or edit my question? $\endgroup$
    – Bibek Ghimire
    Jul 4, 2018 at 2:52
  • $\begingroup$ Edit your question. $\endgroup$
    – Christian Blatter
    Jul 4, 2018 at 8:15
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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.

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  • $\begingroup$ according to you one of the 5- digit number is 11112, am i right? $\endgroup$
    – Bibek Ghimire
    Jun 29, 2018 at 16:09
  • $\begingroup$ The OP says you must each digit at least once. Why not give just a hint? $\endgroup$
    – Ethan Bolker
    Jun 29, 2018 at 16:10
  • $\begingroup$ @BibekGhimire Yes sorry I didn't notice that part. $\endgroup$
    – user
    Jun 29, 2018 at 16:17
  • $\begingroup$ you have to form 5 digit even number and all the digits 1,2,3 and 6 must be used, please read the problem carefully. @Ethan $\endgroup$
    – Bibek Ghimire
    Jun 29, 2018 at 16:17
  • $\begingroup$ @EthanBolker Yes I lost that part, now it should be fixed. $\endgroup$
    – user
    Jun 29, 2018 at 16:18

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