360
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 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."
There are 12 two letter arrangements of the letters in PARK.
The word "math" consists of 4 distinct letters: m, a, t, and h. To find the number of three-letter arrangements, we can use the permutation formula for selecting and arranging 3 letters from 4 distinct letters, which is given by ( P(n, r) = \frac{n!}{(n-r)!} ). Here, ( n = 4 ) and ( r = 3 ), so the calculation is ( P(4, 3) = \frac{4!}{(4-3)!} = \frac{4!}{1!} = 4 \times 3 \times 2 = 24 ). Thus, there are 24 different three-letter arrangements.
There are 172 different arrangements.
The number of different three letter arrangements that can be done from theletters in the word "mathematics"is; 11P3 =11!/(11-3)! =990
To find the number of distinct 4-letter words that can be formed from the letters in "aaggre," we first note the available letters: a, a, g, g, r, e. The distinct combinations of letters can vary depending on how many times each letter is used. For instance, if we select two 'a's, we can combine them with different arrangements of 'g's, 'r', and 'e' to form words. The total number of distinct 4-letter combinations can be calculated by considering the different cases based on the repetitions of letters, leading to a total of 30 unique 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.
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 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 "candy" consists of 5 distinct letters: c, a, n, d, and y. To find the number of different two-letter arrangements, we can choose 2 letters from the 5 and then arrange them. The number of ways to choose 2 letters from 5 is given by the combination ( \binom{5}{2} = 10 ), and since each pair can be arranged in ( 2! = 2 ) ways, the total number of two-letter arrangements is ( 10 \times 2 = 20 ). Thus, there are 20 different two-letter arrangements.
There are 12 two letter arrangements of the letters in PARK.
The word "math" consists of 4 distinct letters: m, a, t, and h. To find the number of three-letter arrangements, we can use the permutation formula for selecting and arranging 3 letters from 4 distinct letters, which is given by ( P(n, r) = \frac{n!}{(n-r)!} ). Here, ( n = 4 ) and ( r = 3 ), so the calculation is ( P(4, 3) = \frac{4!}{(4-3)!} = \frac{4!}{1!} = 4 \times 3 \times 2 = 24 ). Thus, there are 24 different three-letter arrangements.
There are 172 different arrangements.
Eider
fluffy
fuzzy