6 digit Numbers from 4 digits

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The total number of permutations without the constraint that the permutation should have all digits present is: 4^6

Let us take the complementary set:

Consisting of exactly one digit: 4 Consisting of exactly two digits: (4,2) * (2^6-2) Consisting of exactly three digits: (4,3) * (3^6-3 * (2^6-2) -3)

Hence the answer is:

4^6 - 4 - (4,2) * (2^6-2) - (4,3) * (3^6-3 * (2^6-2) -3) = 1560

Consider the first four numbers. There are only 3 possibilities:

1. There are four different numbers
2. There are three different numbers
3. There are two different numbers

Any less would mean there could not be four numbers total.

Now simply add up the totals for each possibility:

1. There are 4! ways for the first four numbers, and 4 ways for each the 5th and 6th, so $24*4*4=384$
2. There are 4 possibilities for the number left out, 3 possibilities for the number repeated, and 12 possibilities for the positioning of the two singles among the first four, for a total of $144$ among the first four. Among the last two, there are 7 possibilities (1 if both are the same as we still need to use up one number, and 6 if they aren't, as there are 2 places for the forced number, and 3 choices for the other.) $7*144=1008$
3. There are 6 possibilities for the 2 to choose and 14 ways to put them in the first 4 slots ( $2*2*2*2$ minus the 2 ways which only contain one number). There are two permutations of the last 2 digits, for a total of $6*14*2=168$

$384+1008+168=\boxed{1560}$

There are two cases here. First, there is a digit that is repeated three times and the other three digits are written once in a number. Second, there are two digits which is repeated twice and the other two digits are repeated just once in a number.

First case: Let one of the number is $~~111234$

Each of digit can be repeated maximum three times in a number, so there is $\left(\begin{matrix}4\\1\end{matrix}\right)=4$ ways to select the digit that is repeated three times in a number. Thus, we can form $4\times{\frac{6!}{3!\times{1!}\times{1!}\times{1!}}}=480$ numbers.

Second case: Let one of the umber is $~~334412$

There are two digits can be repeated twice on a number, so there is $\left(\begin{matrix}4\\2\end{matrix}\right)=6$ ways to select the digit that is repeated twice in a number. So, we have $6\times{\frac{6!}{2!\times{2!}\times{1!}\times{1!}}}=1080$ numbers.

Thus, the total number can be formed is $480+1080=1560$ numbers.