That's eight letters, so: 8! = 40320 different arrangements.
n! means "factorial", and the expression expands to n*(n - 1)*(n - 2) ... * 2 * 1
The number of 7 letter permutations of the word ALGEBRA is the same as the number of permutation of 7 things taken 7 at a time, which is 5040. However, since the letter A is duplicated once, you have to divide by 2 in order to find out the number of distinct permutations, which is 2520.
Answer 9!/3!2!=30240 Note there are three a's and two p's and a total of 9 letters in apparatus. So there are 9 choices for the first letter, and 8 for the next, then 7 for the third etc. The we divide by 3! and 2! to avoid overcounting. The answer is 9!/3!2! The numerator can be written as 9! which is pronounced 9 factorial. Then we divide that by 3! times 2! In general if you have have n letters, there are n! words you can make, However, if some of the letters are repeated you must divide to avoid overcounting. If letter a is repeated r1 times, b r2 times, c r3 times.. etc, Then the number of words is n!/r1!r2!r3!
The letters T and L are formed as asked.l and t from fiona1234
X and T because the they can still make the T shape and still be perpendicular
In a set, as it is usually defined, elements can't be repeated. "Mathematics" has 8 distinct letters, so your set would have 8 letters. The number of possible subsets (this includes the empty set, and the set itself) is two to the power 8.
240
Algorithm is the only nine letter word.
120
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.
720 (6 x 5 x 4 x 3 x 2)
If the first letter must be either W or K, there are 2 choices for the first letter. For the remaining 4 letters, since repeats are allowed, each can be any of the 26 letters in the alphabet. Therefore, the total number of arrangements is (2 \times 26^4), which equals (2 \times 456976 = 913952).
8 different 4-letter words can be formed from the letters of the word "Nation".
There are 7!/(3!*2!) = 420 ways.7! for the seven letters in "success", butthere are 3 s which are indistinguishable, so divide by 3!there are 2 c which are indistinguishable, so divide by 2!
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.
Words that can be made with the letters in 'gazebo' are:aageagobagbebegboabogegogabgazegogob
The word "IMAGINARY" has 9 letters, including 4 vowels (I, A, I, A) and 5 consonants (M, G, N, R, Y). To find the number of arrangements where the vowels do not come together, first calculate the total arrangements of the letters (considering the repeated vowels) and then subtract the arrangements where the vowels are together, treating them as a single unit. After performing these calculations, the total number of arrangements where the vowels never come together is found to be 2880.
There are a total of 15 letters in "season greetings." To calculate the number of words that can be formed, we first need to determine the number of unique arrangements of these letters. This can be calculated using the formula for permutations of a multiset, which is 15! / (2! * 2! * 2! * 2! * 2! * 2! * 1!). This results in 1,816,214,400 unique arrangements. However, not all of these arrangements will form valid English words, as many will be nonsensical combinations of letters.