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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 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 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.
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.
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Since you are using in the arrengement the all 4 letters, then there are 4! = 4*3*2*1 = 24 permutations.
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 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.
The word MATHEMATICS has 11 letters. The number of permutations of 11 things taken 11 at a time is 11 factorial (11!), or 39,916,800.
well let's see....setetstseeststetesuh.... that would be 6Mathematically, it would be the number of permutations of three things taken three at a time=3!=3*2*1=6.
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 n objects taken all together is n!.
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.
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.
Permutations of 10 letters taken 3 a time = 10 x 9 x 8 = 720