# Permutations with repetitions

How many times can the input of 1.2.2.3.3.3.4 be permutated into four digits, three digits, and two digits without repetition?
Ex:
4 digits = 1223, 2213, 3122, 2313, 4321. . etc
3 digits = 122.212.213.432. . etc
2 digits = 12, 21, 31, 23

I have tried the permutation formula, and it failed.

n4 =  115
n3 =  43
n2 =  14

## Step-by-step explanation:

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Danka
CAN YOU BE MORE SPECIFIC WITH THE ANSWERS?

Math student
Thats wrong:

For anyone wondering why you got the wrong answer, here is the real explanation:

To find the number of permutations without repetition for a given set of numbers, we can use the formula for permutations:

P(n, r) = n! / (n - r)!

Where P(n, r) denotes the number of permutations of n items taken r at a time, and n! represents the factorial of n.

Let's calculate the number of permutations for the input 1.2.2.3.3.3.4.

For four digits (r = 4):
n = 7 (total number of digits)
r = 4 (number of digits to be taken at a time)

P(7, 4) = 7! / (7 - 4)!
= 7! / 3!
= (7 * 6 * 5 * 4 * 3 * 2 * 1) / (3 * 2 * 1)
= 7 * 6 * 5 * 4
= 840

So, there are 840 permutations of four digits without repetition.

For three digits (r = 3):
n = 7 (total number of digits)
r = 3 (number of digits to be taken at a time)

P(7, 3) = 7! / (7 - 3)!
= 7! / 4!
= (7 * 6 * 5 * 4 * 3 * 2 * 1) / (4 * 3 * 2 * 1)
= 7 * 6 * 5
= 210

So, there are 210 permutations of three digits without repetition.

For two digits (r = 2):
n = 7 (total number of digits)
r = 2 (number of digits to be taken at a time)

P(7, 2) = 7! / (7 - 2)!
= 7! / 5!
= (7 * 6 * 5 * 4 * 3 * 2 * 1) / (5 * 4 * 3 * 2 * 1)
= 7 * 6
= 42

So, there are 42 permutations of two digits without repetition.

11 months ago  1 Like

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