There are 34650 distinct orders.There are 34650 distinct orders.There are 34650 distinct orders.There are 34650 distinct orders.
The word "BOX" consists of 3 distinct letters. The number of arrangements of these letters can be calculated using the factorial of the number of letters, which is 3! (3 factorial). Therefore, the total number of arrangements is 3! = 3 × 2 × 1 = 6. Thus, there are 6 possible arrangements of the letters in "BOX."
The word "spineless" has 9 letters, including 3 s's and 2 e's, so the number of distinct permutations of the letters is: 9!/(3!2!) = 30,240
There are 12 two letter arrangements of the letters in PARK.
If all the letters are unique in the set, there are 6 choices for the first letter, 5 for the second letter, 4 for the third letter, etc. This results in 6 X 5 X 4 X 3 X 2 = 720 arrangements. If some of the six letters are duplicated, there will be fewer distinct arrangements.
from itertools import permutationsfrom string import joinarrangements = [ ]for p in permutations ( 'ugccsir', 6 ) :arrangement = join ( p, '' )if 'ui' in arrangement :continueif not arrangement in arrangements :arrangements . append ( arrangement )print len ( arrangements )The above Python code gives a result of 2,220.
There are 6!/(3!*2!) = 60 arrangements.
There are 34650 distinct orders.There are 34650 distinct orders.There are 34650 distinct orders.There are 34650 distinct orders.
The number of distinct arrangements of the letters of the word BOXING is the same as the number of permutations of 6 things taken 6 at a time. This is 6 factorial, which is 720. Since there are no duplicated letters in the word, there is no need to divide by any factor.
Take note of the word "surprising":There are 10 letters total.There are 2 r's.There are 2 i'sThere are 2 s's.There are 10! total ways to arrange the letters. Since repetition is not allowed for the arrangements, we need to divide the total number of arrangements by 2!2!2! Therefore, you should get 10!/(2!2!2!) distinct arrangements
The word "BOX" consists of 3 distinct letters. The number of arrangements of these letters can be calculated using the factorial of the number of letters, which is 3! (3 factorial). Therefore, the total number of arrangements is 3! = 3 × 2 × 1 = 6. Thus, there are 6 possible arrangements of the letters in "BOX."
The word "spineless" has 9 letters, including 3 s's and 2 e's, so the number of distinct permutations of the letters is: 9!/(3!2!) = 30,240
The number of different three letter arrangements that can be done from theletters in the word "mathematics"is; 11P3 =11!/(11-3)! =990
There are 12 two letter arrangements of the letters in PARK.
If all the letters are unique in the set, there are 6 choices for the first letter, 5 for the second letter, 4 for the third letter, etc. This results in 6 X 5 X 4 X 3 X 2 = 720 arrangements. If some of the six letters are duplicated, there will be fewer distinct arrangements.
The number of 5 letter arrangements of the letters in the word DANNY is the same as the number of permutations of 5 things taken 5 at a time, which is 120. However, since the letter N is repeated once, the number of distinct permutations is one half of that, or 60.
The number of permutations of the letters in the word SCHOOLS is the number of permutations of 7 things taken 7 at a time, which is 5040. However, since two of the letters, S and O, are duplicated, the number of distinct permutations is one fourth of that, or 1260.