Find the number of six-digit numbers that can be formed

Question

find the number of six-digit numbers that can be formed using the digits from the number 112 233. If these numbers are arranged in ascending order,find a.) the largest number. b.) the 30th number. c.)the median .

Here is where i get stuck. for a, i guess the answer is 33 2211.

for b.) 1x1x 4! =24 means that 6 more lest form 30 th so i managed to list out the outcome: 121233, 122133, 122313 122331 132231 132331. so , my answer of b is 132331. but the answer given is 133221. for c, it is more worse. i have no idea about it.

Answer 1

To find the number of such six-digit numbers we can use the multinomial coefficient for the permutations of $6$ objects given that $2$ are equal (the $1$'s), another $2$ are equal (the $2$'s), and another $2$ are equal (the $3$'s):

$${6 \choose 2,2,2}=\frac{6!}{2!2!2!}=\frac{720}{2\cdot 2\cdot 2}=90$$

If you are not familiar with multinomial coefficients, think of the number of ways of arranging all six digits if there were equal: that is $6!=720$. But the two $1$'s could be swapped, dividing that count by $2$, and similarly for the $2$'s and the $3$'s. This gives the same count.

For (a), the largest number, we make a number the largest by putting the largest digits at the front and the smallest digits at the back. That gives us

$$332211$$

For (b), we saw there are $90$ numbers total, and we also see that one-third of them start with $1$, one-third start with $2$, and one-third start with $3$. That means the first $30$ numbers start with $1$, so the $30$'th number is the largest (last) one that starts with $1$. So again we put the large digits to the front but after the initial $1$, giving us

$$133221$$

For (c), for each number in the list we can get another one by changing all the $1$'s to $3$'s and all the $3$'s to $1$'s. If we do that for each number in the list we reverse the order of the list. Note that the $2$'s just stay the same. So the list revolves around the number made of all $2$'s, which is therefore the median of the list:

$$222222$$

The $45$'th number in the list, the first of the two middle numbers, is $221331$. The $46$'th number in the list, the second of the two middle numbers, is $223113$. The median is the average of those two numbers, again giving us $222222$.