My question is, if I have for example digits $\{1,2,3\}$, how many $4$ digit numbers can I obtain if I have to use all of them in each combination?
Correct combinations:
$\{1,1,2,3\}$
$\{1,1,3,2\}$
$\{3,2,1,2\}$
$\{2,2,3,1\}$
Wrong combinations:
$\{1,1,2,1\}$ There is not digit $\{3\}$
$\{1,1,3,3\}$ There is not digit $\{2\}$
$\{2,2,2,3\}$ There is not digit $\{1\}$
And which would be the probability of this happening? I guess that finding the combinations all I have to do is to divide this result by the total number of possible combinations.
$\endgroup$4 Answers
$\begingroup${1,2,3,A} 4 digits combinations: 4x3x2x1=24
{1,2,3,1} 4 digits combinations: 4x3x2x1/2=12 since 1,1 replacing 1,A A,1
{1,2,3,1} 4 digits combinations: 12
{1,2,3,2} 4 digits combinations: 12
{1,2,3,3} 4 digits combinations: 12
Total: 12x3=36
Hope it helps
$\endgroup$ 1 $\begingroup$Now for $2$ digits in $4$ places we first put the $2$ digits in any $2$ places in only $1$ way( since we have to place both of them).Now there are $2$ places which are still empty.These two places can be filled in two different ways
$(1)$ We can place any $1$ digit twice like $\{1222\}\{1211\}...$
$(2)$ We can place $2$ digits at the $2$ places like $\{1212\}\{1221\}$ and so on.
For $(1)$ the number of such numbers is $2 \cdot (4!/3!)=8$ (you have calculated) For $(2)$ the number of such numbers is $4!/(2! \cdot 2!) =6$
So adding $(1)$ and $(2)$ we get the total number of such numbers i.e $8+6=14$ So the probability is $14/2^4=0.875$.
So you should be very careful how we can fill those empty places after placing the digits. While calculating you need to consider all possible ways of filling those empty places(here you missed out the possibility $(2)$). Hope you have got your answer.
$\endgroup$ 2 $\begingroup$Hint: First, pick the digit to be doubled-how many ways? Then pick the slots it goes in-how many ways? Then pick the ways to put the other two digits in the remaining slots. For the probability, divide by the number of four digit numbers you can create from this list of digits-how many is that?
$\endgroup$ $\begingroup$If the $3$ digits are specified:
The problem solving strategy is as follows:
At first we will consider how we can place those $3$ digits in $4$ places and then we will consider how we can arrange them in those places. We have to place all the $3$ digits in each combination.So it can be done in only $1$ way.
Now we see that after placing the $3$ digits $1$ place is still empty.This empty place can be filled in $3$ ways( any of the $3$ digits can be placed there).
Now we have to arrange them.So, the number of arrangements is $4!/2!$(since we have $4$ digits of which $2$ digits are same).So, number of $4$ digit numbers $= 3 \cdot (4!/2!)=36 $.
Total number of $4$ digit numbers using those $3$ digits is $3^4$. So the probability is $36/3^4= 4/9$.
