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How many different $4$-digit numbers can be made from digits of number $426269$ with given rules if every digit can appear the number of times it appears in the number $426269$ ($2 \times 2, 2 \times 6, 4, 9$)?

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This is a multinomial distribution. Permute the digits, then divide out by the symmetry groups of each element:

$$ \dfrac{6!}{2! * 2!}$$

So since $2$ appears twice, switching the orders of the two $2$ characters will not change the given string. Hence, we divide out. The same applies with the $6$ characters.

Edit: Since you edited for $4$-digit numbers, I will update. So we consider a few cases. The first case is that all digits are distinct. There are four distinct digits, so there are $4!$ possible combinations.

If we have two $2$ characters, we use a multinomial distribution $\dfrac{4!}{2!}$ to divide out by the symmetry group. However, we must select two elements from the three distinct elements. So we have $\dfrac{4! * 3}{2!}$ possibilities. We get the same number of elements when using both $6$ elements, so we multiply by $2$ to get $4! * 3$ possibilities.

There are $\dfrac{4!}{2! * 2!} = 3!$ possible strings using only the digits $2, 6$. And so since these quantities are disjoint, add them up: $4! + 3 * 4! + 3! = 102$.

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I get $180$ numbers. Note that this assumes that each digit appears exactly as many times as specified. – ml0105 Apr 27 '14 at 21:41
excuse me, I missed one detail when asking this question... It is about 4 digit numbers. – slimDeviant Apr 27 '14 at 21:41
I've updated my answer. – ml0105 Apr 27 '14 at 21:45
That was my way of reasoning this problem too and the same result but my textbook says that result is supposed to be 104. What do you think about that? – slimDeviant Apr 27 '14 at 21:53
I forgot to choose two of the three characters for the second case. But I got 102, not 104. – ml0105 Apr 27 '14 at 21:59

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