# 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. Apr 27, 2014 at 21:41
• excuse me, I missed one detail when asking this question... It is about 4 digit numbers. Apr 27, 2014 at 21:41
• I've updated my answer. Apr 27, 2014 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? Apr 27, 2014 at 21:53
• I forgot to choose two of the three characters for the second case. But I got 102, not 104. Apr 27, 2014 at 21:59