Q: Does the number of permutations always exceed the number of combinations?

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Yes

There are 120 permutations and 5 combinations.

On a calculator: 43C5 = 962598 This is for combinations and not permutations, so in essence, the order of the 5 number combinations does not matter. Yours Truly, Mr Greatness

The number of combinations - not to be confused with the number of permutations - is 2*21 = 42.

There are only four combinations but there are 8 permutations.

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Yes

There are 120 permutations and 5 combinations.

On a calculator: 43C5 = 962598 This is for combinations and not permutations, so in essence, the order of the 5 number combinations does not matter. Yours Truly, Mr Greatness

The number of combinations - not to be confused with the number of permutations - is 2*21 = 42.

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.

Not quite. Number of combinations is 20, number of permutations is 10. Any 2 from 5 is 10 but in any order doubles this.

Assuming that "number" means digits and that it is permutations (rather than combinations) that are required, the answer is 816 which is 281.475 trillion (approx).

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

There are only four combinations but there are 8 permutations.

It depends on how many digit you are choosing from.

All the numbers from 0 to 999 so 1000 numbers However if you mean only using each number once the answer is about 720 * * * * * No. The above answer refers to the number of permutations. Permutations are NOT the same as combinations as anyone who has studies any probability theory can tell you. The number of combinations is 9C3 = 9*8*7/3*2*1 = 84

120 WRONG! That is the number of PERMUTATIONS. In the case of combinations, the order of the numbers does not matter, so there is only 1 5-number combination from 5 numbers.