Q: Is the number of permutations of two items from a data set is always two times the number of combinations when taking two objects at a time from the same data set?

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No. The number of permutations or combinations of 0 objects out of n is always 1. The number of permutations or combinations of 1 object out of n is always n. Otherwise, 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.

If there are n different objects, the number of permutations is factorial n which is also written as n! and is equal to 1*2*3*...*(n-1)*n.

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No. The number of permutations or combinations of 0 objects out of n is always 1. The number of permutations or combinations of 1 object out of n is always n. Otherwise, yes.

If there are n objects and you have to choose r objects then the number of permutations is (n!)/((n-r)!). For circular permutations if you have n objects then the number of circular permutations is (n-1)!

The number of permutations of n objects taken all together is n!.

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

Yes

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.

If there are n different objects, the number of permutations is factorial n which is also written as n! and is equal to 1*2*3*...*(n-1)*n.

The formula for finding the number of distinguishable permutations is: N! -------------------- (n1!)(n2!)...(nk!) where N is the amount of objects, k of which are unique.

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).