Given a four digit number XXXX, are there more permutations if repetition is allowed or not allowed?
with no repetitions, there are only 4! permutations.
With repetition, there are 10^4 (or maybe 9*10^3 if you disallow leading zeros) choices.
if a zero is allowed in the lead position, then with no repetition, there would be
10(9)(8)(7) ways
if a zero is allowed in the lead position, and repetition is allowed, there would be
10^4 ways
If zero is not allowed, with no repetition: 9(9)(8)(7)
zero not allowed, but with repetition: 9(10^3)
If XXXX is a specified number, and the numbers are all different,
then 4! , as Steve said in his first answer
If XXXX is a specified number and some of the numbers are duplicates, then more
information is needed.
Oh, well that's an interesting question. You see, when repetition is allowed, it's like having a buffet with endless options. It's like going to a pizza place where you can have as many toppings as you want, even extra cheese! You can mix and match the digits however you please, resulting in a larger number of permutations.
But when repetition is not allowed, it's like going to a fancy restaurant with a fixed menu. You're limited to the choices they offer, and you can't have the same digit more than once. So, in this case, the number of permutations would be reduced because you can't repeat any digits.
So, in a nutshell, repetition allows for more "flavorful" permutations, while without repetition, it's like having a more limited menu.
To determine the number of permutations with repetition allowed and not allowed for a four-digit number XXXX, we can use the concept of permutations.
1. Permutations with repetition allowed:
When repetition is allowed, each digit can take any value from 0 to 9. Therefore, each digit has 10 options (0-9). So, the total number of permutations with repetition allowed is calculated as 10^4 (since we have 4 digits).
2. Permutations with repetition not allowed:
When repetition is not allowed, the first digit can take any value from 1 to 9 (as leading zeros are not allowed). So, there are 9 options for the first digit.
For the second digit, there are only 9 remaining options, as one digit is already used (repetition not allowed). Similarly, for the third digit, there are 8 remaining options, and for the fourth digit, there are 7 remaining options.
Therefore, the total number of permutations with repetition not allowed is calculated as 9 * 9 * 8 * 7 = 4,536.
Comparing the two scenarios, we find that there are more permutations when repetition is allowed (10,000 permutations) compared to when repetition is not allowed (4,536 permutations).
To determine whether there are more permutations with repetition allowed or not allowed for a four-digit number XXXX, we need to calculate the number of permutations in each case.
When repetition is allowed, any digit (0-9) can be chosen for each of the four positions independently. Therefore, we have 10 options for each position. So, the total number of permutations with repetition allowed is calculated as:
10 options for the first position × 10 options for the second position × 10 options for the third position × 10 options for the fourth position = 10^4 = 10,000 permutations.
On the other hand, when repetition is not allowed, we have to ensure that each digit chosen is unique for each position. For the first position, we have 10 options to choose from (0-9). For the second position, we have 9 options left (as one digit has already been chosen). For the third position, we have 8 options left, and for the fourth position, we have 7 options left.
Therefore, the total number of permutations without repetition allowed is calculated as:
10 options for the first position × 9 options for the second position × 8 options for the third position × 7 options for the fourth position = 10 × 9 × 8 × 7 = 5,040 permutations.
Comparing both results, we can see that there are more permutations when repetition is allowed (10,000 permutations) compared to when repetition is not allowed (5,040 permutations).