If these numbers(112233)are arranged to form six-digit numbers in ascending order,how to find the 30th number? and the median?

If we were dealing with 6 unique numbers, there'd be #6! = 720# ways to arrange them. However, we have to divide out the duplicates from the repeated numbers (and we do that by dividing by the number of ways we can internally order the repeated numbers, which is #2!# for each of the 1s, 2s, and 3s, so #(2!)^3 = 8#. This gives:

#720/6=90# different ways we can order the numbers.

We can divide these numbers into 30 (1-30) that start with the number 1, 30 (31-60) that start with the number 2, and 30 (61-90) that start with the number 3.

The 30th number therefore is the biggest number we can make starting with a 1, giving

133221

The median is the middle term. When dealing with an even number of terms, we take the 2 middle terms and take the mean. In this case, the two terms in question are the 45th and 46th terms. Let's find them.

We know that 30 numbers, 31-60, start with 2. We can divide up the list again by the number in the second position. There are 5 numbers that can be in the second position (two of the 1, one of the 2, and two of the 3), so we can divide the list of 30 numbers into chunks of six each.

21..., 31-36
21..., 37-42
22..., 43-48
23..., 49-54
23..., 55-60

The 45th and 46th terms are in the 22... group:

221133, 43th
221313, 44th
221331, 45th
223113, 46th
223131, 47th
223311, 48th

And so we take the mean of #221331# and #223113#, which is:

#(221331+223113)/2=444444/2=222222#