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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!

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    what do you mean the threes and fours are not in the beginning? – 2017-02-21
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    sounds like all you mean is $\binom 74 -2=33$, no? – 2017-02-21
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    @lulu I agree.. – 2017-02-21
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    My thinking: ignoring the constraint, you just chose the four slots to put the $4's$, that's $\binom 74$. then you subtract the $2$ cases you don't like. Or did you mean something else? – 2017-02-21
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    I just have the set {3,3,3,4,4,4,4} with this set I have to create digits of the length seven. There is no repetition. 4444333 for example is not allowed because 4444 are the first digits, 3344443 or 4443333 would be okay. The same for the threes – 2017-02-21
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    You responded with a comment a full ten minutes after lulu's comment. Have you yet attempted to understand lulu's comment? What do or don't you understand about it? Do you understand why there are $\binom{7}{4}$ arrangements of three 3's and four 4's? Do you understand why exactly two of those are "bad" and should be removed from our count? Note: $7!$ is the number of seven digit numbers created from $\{1,2,3,\dots,7\}$, not the number of seven digit numbers created from $\{3,3,3,4,4,4,4\}$. – 2017-02-21

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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.

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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.