September has 9 letter, of which one appears 3 times.
So the number of distinct permutations is 9!/3! = 120,960
Take the total number of letters factorial, then divide by the multiple letters factorial (a and e). 7! / (2!*2!) or 1260.
There are ten letters in the word. The total number of possible permutations is(10) x (9) x (8) x (7) x (6) x (5) x (4) x (3) x (2) = 3,628,800But the two 'c's can be arranged in either of 2 ways with no distinguishable change.Also, the three 'i's can be arranged in any of (3 x 2) = 6 ways with no distinguishable change.And the three 't's can be arranged in any of (3 x 2) = 6 ways with no distinguishable change.So the total number of possible permutations can be divided by (2 x 6 x 6) = 72, the number oftimes each distinguishable permutation occurs with different and indisnguishable arrangementsof 'c', 'i', and 't'.We're left with(10) x (9) x (8) x (7) x (...) x (5) x (...) x (...) x (2) = (3,628,800/72) = 50,400 distinguishable arrangements.
The distinguishable permutations are the total permutations divided by the product of the factorial of the count of each letter. So: 9!/(2!*2!*1*1*1*1*1) = 362880/4 = 90,720
Permutations are the different arrangements of any number of objects. When we arrange some objects in different orders, we obtain different permutations.Therefore, you can't say "What is the permutation of 5?". To calculate permutations, one has to get the following details:The total number of objects (n) (necessary)The number of objects taken at a time (r) (necessary)Any special conditions mentioned in the question (optional).
Permutation Formula A formula for the number of possible permutations of k objects from a set of n. This is usually written nPk . Formula:Example:How many ways can 4 students from a group of 15 be lined up for a photograph? Answer: There are 15P4 possible permutations of 4 students from a group of 15. different lineups
The solution is count the number of letters in the word and divide by the number of permutations of the repeated letters; 7!/3! = 840.
permutation without replacement
The word "numbers" consists of 7 distinct letters. The number of permutations of these letters is calculated using the factorial of the number of letters, which is 7!. Therefore, the total number of permutations is 7! = 5,040.
Take the total number of letters factorial, then divide by the multiple letters factorial (a and e). 7! / (2!*2!) or 1260.
The word mathematics has 11 letters; 2 are m, a, t. The number of distinguishable permutations is 11!/(2!2!2!) = 39916800/8 = 4989600.
There are ten letters in the word. The total number of possible permutations is(10) x (9) x (8) x (7) x (6) x (5) x (4) x (3) x (2) = 3,628,800But the two 'c's can be arranged in either of 2 ways with no distinguishable change.Also, the three 'i's can be arranged in any of (3 x 2) = 6 ways with no distinguishable change.And the three 't's can be arranged in any of (3 x 2) = 6 ways with no distinguishable change.So the total number of possible permutations can be divided by (2 x 6 x 6) = 72, the number oftimes each distinguishable permutation occurs with different and indisnguishable arrangementsof 'c', 'i', and 't'.We're left with(10) x (9) x (8) x (7) x (...) x (5) x (...) x (...) x (2) = (3,628,800/72) = 50,400 distinguishable arrangements.
It is 6! = 6*5*4*3*2*1 = 720
There can be only one permutation of a single number: so the answer is 7.
First of all, find the total number of not-necessarily distinguishable permutations. There are 12 letters in hippopotamus, so use 12! (12 factorial), which is equal to 12 x 11x 10 x9 x8 x7 x6 x5 x4 x3 x2 x1. 12! = 479001600.Then count the of each letter and calculate how many permutations of each letter can be made. For example, here is 1 h, so there is 1 permutation of 1 h.H 1I 1P 60 2T 1A 1M 1U 1S 1Multiply these numbers together. 1 x1 x6 x2 x1 x1 x1 x1 x1 = 12Divide 12! by this number. 479001600 / 12 = 39,916,800 Distinguishable Permutations.
We can clearly observe that the word "ellises" has 7 letters and three pairs of letters are getting repeated that are 'e','l' and 's'. So, Number of distinguishable permutations = 7!/(2!2!2!) = 7 x 6 x 5 x 3 = 630.
nPrwhere:n number of objectsr is number of arrangements
For the first letter, you can have any of 5 choices. Then for the next, you can have any of four. Then for the next, three and so on. Thus the number of permutations can be calculated by 5x4x3x2x1. Doing this gives 120. Therefore the number of permutations of the letters in the word smart is 120.