Combinatorics on numbers with given digits I have this combinatorics problem:
How many numbers with seven digits can be created with three times digit 3 and four times digit 4, under the condition that the threes and the fours are not in the beginning. So 4444333 or 3334444 are not allowed.
My ideas:
I can totally create 7! numbers. 
Then I have to subtract these numbers in the beginning, so 7!-(4!+3!)=5010
Am I right with my thoughts?
Thx in advance!
 A: You are using permutations on your reasoning but without taking into account the repetition, since for a normal permutation all objects have to be different, this is not the case ($3$ repeats $3$ times, $4$ repeats $4$ times).
Then you have to define a permutation of that set but considering the repetition of $3's$ and $4's$. That is, a permutation of multiset.
Formula is: $PR^{7}_{4,3} = \frac{7!}{4!3!}=35$ but since $4's$ and $3's$ cannot appear all together in the prefix nor the suffix then we have to exclude $2$ possibilities, then the answer is $33$. 
As in my country (Spain), we use $PR$, and we call this a permutation with repetition, but in standard combinatorics it is called permutation of multiset.
A: One digit $3$ is identical to another digit $3$, so there is only one ordering of a string $33$ for example.
What you need is to choose the digit positions occupied by one of the digits (say the $3$s) leaving the other positions to be filled by the other digit. 
This is given by the binomial coefficient $\binom {\large 7}{\large 3} = \frac{\large 7!}{\large 3!(7-3)!}$, derived by chosing the positions $7\cdot 6\cdot 5 = \frac{\large 7!}{\large (7-3)!}$ ways and then dividing by $3!$ also to account for choosing the same locations in a different order.
As you can probably recognize, $\binom 73 = \binom 74$, since $4=7-3$, so we would get the same result by choosing positions for the $4$ digits.
