how many numbers of six digits can b made by arranging the digits of 123425 so that all the even digits do not occpy consecutive places?

To solve the problem of how many six-digit numbers can be formed by arranging the digits of 123425 such that all even digits do not occupy consecutive places, we first need to analyze the digits involved. The digits we have are 1, 2, 2, 3, 4, and 5. In this set, the even digits are 2, 2, and 4. We want to ensure that the even digits are not next to each other in any arrangement.

Step-by-Step Breakdown

Let’s start by identifying the total number of arrangements of the digits without any restrictions. Since the digit 2 appears twice, we will use the formula for permutations of multiset:

The total arrangements can be calculated using the formula:

n! / (n1! * n2! * ... * nk!)

Here, n is the total number of items to arrange, and n1, n2, etc., are the counts of each distinct item. For our digits:

• Total digits (n) = 6 (1, 2, 2, 3, 4, 5)
• Count of 2 = 2 (since it appears twice)

Thus, the total arrangements are:

6! / (2!) = 720 / 2 = 360

Placing the Odd Digits

Next, we focus on arranging the digits such that the even digits (2, 2, 4) are not adjacent. First, let's arrange the odd digits, which are 1, 3, and 5. The arrangement of the odd digits can be calculated as follows:

Since all odd digits are distinct, their arrangements are:

3! = 6

When we arrange the odd digits, we create gaps where the even digits can be placed. For example, if we arrange the odd digits (1, 3, 5), we can visualize it like this:

• _ 1 _ 3 _ 5 _

In this arrangement, there are 4 gaps (before the first digit, between digits, and after the last digit) where we can place the even digits. We must select 3 out of these 4 gaps to place the even digits while ensuring that they do not occupy consecutive positions.

Choosing Gaps for Even Digits

We need to choose 3 gaps from the 4 available ones. The number of ways to choose 3 gaps out of 4 can be calculated using the combination formula:

C(n, k) = n! / (k!(n-k)!)

So, we have:

C(4, 3) = 4! / (3! * 1!) = 4

Arranging the Even Digits

Now, we need to arrange the even digits (2, 2, 4) in the selected gaps. The arrangements of the even digits, considering the repetition of the digit 2, can be calculated similarly:

3! / (2!) = 6 / 2 = 3

Final Calculation

Now, we can combine all of these calculations to find the total number of arrangements where even digits do not occupy consecutive places:

Total Arrangements = (Arrangements of Odd Digits) * (Ways to Choose Gaps) * (Arrangements of Even Digits)

Total Arrangements = 6 * 4 * 3 = 72

So, the final answer is that there are 72 different six-digit numbers that can be formed by arranging the digits of 123425 such that the even digits do not occupy consecutive places.