The answer is number 5
F, I, V, E= 4 letters
But, still has 5 when the 3 letters-"F", "I", "E" are removed because "V" stands for "5" as in the Roman numbers.
-- any mixed number -- any mis-spelled bagel or donut -- the word "wholesome"
The number of different ways you can arrange the letters MNOPQ is the number of permutations of 5 things taken 5 at a time. This is 5 factorial, or 120.
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.
Three-letter words taken from the letters in "careen":acearearccancarneerearecFour-letter words taken from the letters in "careen":careracenearSix-letter words taken from the letters in "careen":careen
No. He may have just not taken it out yet.
Since the word MATH does not have any duplicated letters, the number of permutations of those letters is simply the number of permutations of 4 things taken 4 at a time, or 4 factorial, or 24.
Since there are no duplicate letters in the word RAINBOW, the number of permutations of those letters is simply the number of permutations of 7 things taken 7 at a time, i.e. 7 factorial, which is 5040.
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.
The letters A B O U T can be arranged 120 different ways. This is the number of permutations of 5 things taken 5 at a time, or 5 factorial.
There are 10 letters is the word JOURNALISM. Since they are all different, the number of ways you can arrange them is simply the number of permutations of 10 things taken 10 at a time, or 10 factorial, or 3,628,800.
There is no such word in the Bible, is it spelled correctly?
The number of permutations of the letters in PREALGEBRA is the same as the number of permutations of 10 things taken 10 at a time, which is 3,628,800. However, since the letters R, E, and A, are repeated, R=2, E=2, A=2, you must divide that by 2, and 2, and 2 (for a product of 8) to determine the number of distinctpermutations, which is 453,600.