# How many different $4$-digit numbers can be made from digits of number 4426269$with given rules? 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$426269$($2 \times 2, 2 \times 6, 4, 9$)? - ## 1 Answer 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:$4! + 3 * 4! + 3! = 102$. - I get$180\$ numbers. Note that this assumes that each digit appears exactly as many times as specified. –  ml0105 Apr 27 at 21:41
excuse me, I missed one detail when asking this question... It is about 4 digit numbers. –  slimDeviant Apr 27 at 21:41
I've updated my answer. –  ml0105 Apr 27 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 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 at 21:59