in how many ways could the digits in the number 458 978 be arranged if the prime digits must remain in the original position
i understand that after removing the prime digits you'd be left with 4_8 9_8 and that you have 5 spots where you can put 5 then 4 spots when you can put 7 but after that I am not sure
thats what I had gotten but on my textbook, it says the answer is 24
Well, it seems like you've already made great progress! Now, let's continue with the arrangement.
You have correctly identified that after removing the prime digits, you are left with 4_8 9_8. Now, let's tackle the remaining digits one by one.
First, let's place the digit 5. As you mentioned, there are 5 spots available. Once you've placed the 5, you'll be left with 4_8 9_8.
Next, we move on to the digit 7. There are 4 spots available for this digit, but remember that the digit 5 is already in place. So, we have one less spot to consider. After placing the 7, we have 4_8 9_8.
Now, let's move to the remaining digits: 4, 8, and 9. Since they are non-prime, they can be arranged freely among themselves.
So, to calculate the total number of arrangements, we multiply the possibilities for each digit: 5 (for the digit 5) multiplied by 4 (for the digit 7) multiplied by the remaining arrangements of the non-prime digits (3!). Therefore, the total number of arrangements would be 5 * 4 * 3! (which is 6).
So, the final answer would be 5 * 4 * 3! = 120.
Therefore, there are 120 different ways in which the digits in the number 458 978 can be arranged, with the prime digits remaining in their original positions.
To find the number of ways the digits in the number 458 978 can be arranged if the prime digits (2 and 5) must remain in the original position, you can follow these steps:
1. Identify the prime digits that need to remain in their original position: 2 and 5.
2. Remove the prime digits from the number. After removing 2 and 5, you are left with 4_8 9_8.
3. Now, you have four non-prime digits remaining: 4, 8, 9, and 8. Let's designate them as A, B, C, and D, respectively.
4. Since the prime digits are fixed, the remaining non-prime digits (A, B, C, and D) can be arranged among themselves without any restrictions.
5. There are a total of 4! (4 factorial) ways to arrange the non-prime digits (A, B, C, and D) among themselves. This is because there are four different digits, and factorial notation denotes the product of all positive integers less than or equal to that number. So, 4! = 4 x 3 x 2 x 1 = 24.
6. However, we need to consider that there are repeated digits, 8 appears twice. To avoid counting duplicate arrangements, we need to divide the total arrangements by the number of ways the repeated digits can be arranged among themselves.
7. In this case, there are 2! (2 factorial) ways to arrange the repeated 8s among themselves. So, 2! = 2 x 1 = 2.
8. Now, we can calculate the total number of arrangements by dividing the total arrangements from step 5 by the arrangements of the repeated digits from step 7. Therefore, the number of arrangements is 24 / 2 = 12.
Therefore, there are 12 ways to arrange the digits in the number 458 978 if the prime digits 2 and 5 must remain in the original position.
there are 4!/2! permutations of 4 things with 2 duplicates
well, they can in fact be arranged in 24 ways, but unless you can distinguish the two 8's (say, by color), only 12 will be unique.
Even textbooks can have mistakes.