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.
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.
Correct answer:
Showing 2 comments:
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.
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.
9 months ago 1 Like
Tips for related online calculators
See also our combinations with repetition calculator.
See also our permutations calculator.
See also our variations calculator.
Would you like to compute the count of combinations?
See also our permutations calculator.
See also our variations calculator.
Would you like to compute the count of combinations?
You need to know the following knowledge to solve this word math problem:
Related math problems and questions:
- Repetition 80362
How many six-digit numbers without repetition can be formed from the digits 1, 2, 3, 4, 5, and 6, if the numbers are, to begin with: a) the digit 4; b) digits 4 or 5? - Digits
How many odd four-digit numbers can we create from digits: 0, 3, 5, 6, and 7? (a) the figures may be repeated (b) the digits may not be repeated - Phone numbers
How many 9-digit telephone numbers can be compiled from the digits 0,1,2,..,8,9 that no digit is repeated? - Permutations without repetition
From how many elements can we create 720 permutations without repetition? - How many 13
How many ways can X³ y⁴ z³ be written without an exponent? - Variations 5437
From how many elements can we create six times as many variations of the second class without repetition as variations of the third class without repetition? - Interventions 4321
Draw a target with the given radii. Mark Mirek's interventions and Pepi Mirek's interventions 4,0,3,5,3 Pepa had 14 hits for three shots. - Triangles
Hanka cut the 20 cm long straws into three pieces. Each piece had a length in cm. Then, with these three pieces, she tried to make a triangle. a) What circuit has each of the triangles? b) How long can the longest side measure? c) How many different trian - Variations 4/2
Determine the number of items when the count of variations of the fourth class without repeating is 600 times larger than the count of variations of the second class without repetition. - Salami
We have six kinds of salami that have ten pieces and one kind of salami that has four pieces. How many ways can we distinctly choose five pieces of salami? - Pie II
Vili ate three pieces of pie. If the piece is 1/8, how much pie did he eat? - Divide
How many different ways can three people divide seven pears and five apples? - Four-digit 3912
Create all four-digit numbers from digits 1,2,3,4,5, which can repeat. How many are there? - Chickens and rabbits
In the yard were chickens and rabbits. Together they had 23 heads and 76 legs. How many chickens and how many rabbits were in the yard? - Z9–I–4 MO 2017
Numbers 1, 2, 3, 4, 5, 6, 7, 8, and 9 were prepared for a train journey with three wagons. They wanted to sit out so that three numbers were seated in each carriage and the largest of the three was equal to the sum of the remaining two. The conductor said - Flags
How many different flags can be made from green, white, blue, red, orange, yellow, and purple materials, so each flag consists of three different colors? - Two groups
The group of 10 girls should be divided into two groups with at least four girls in each group. How many ways can this be done?