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Permutations = 4 x 3 x 2 = 24

Combinations = (4 x 3 x 2) / (3 x 2 x 1) = 24/6 = 6

Q: How many permutations and combinations are there of 4 objects taken 3 at a time?

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Just one. * * * * * Depends on how many numbers are on each ring. If there are x numbers, then the total number of combinations (actually they are permutations) is x*x*x or x3.

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 you are using in the arrengement the all 4 letters, then there are 4! = 4*3*2*1 = 24 permutations.

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.

There are 8! = 40320 permutations.

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Do a web search for "permutations and combinations" to find the how. I make it 35,960.

For the permutation the equation is (8!)/(5!) assuming there is no replacement. Which is 8*7*6 = 336 For combinations the equation is (8!)/((5!)(8-5)!) = (8!)/((5!)(3)!) = (8*7*6)/(6) = 56

There are many formules

Trifecta bets are permutations, not combinations, and there are 6840 of them.

720 permutations (the arrangement matters) 120 combinations (the arrangement doesn't matter)

There are 120 permutations and 5 combinations.

8*7/(2*1) = 28

Just 4: 123, 124, 134 and 234. The order of the numbers does not matter with combinations. If it does, then they are permutations, not combinations.

There is only one combination. There are many permutations, though.

It depends on how many digit you are choosing from.

Combinations from 90 objects, taken 3 at a time is 90C3 = 90*89*88/(3*2*1) = 117,480.

There are only four combinations but there are 8 permutations.