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Q: How many permutations can you make with 4 numbers?

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There are 5*4*3 = 60 permutations.

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.

12 numbers

There are 4*3*2*1 = 24 of them.

The number of permutations of the letters SWIMMING is 8 factorial or 40,320. The number of distinct permutations, however, due to the duplication of the letters I and M is a factor of 4 less than that, or 10,080.

Related questions

This is permutations with repetition. The answer is 4^4 = 256 total permutations. Since 2 of the digits used are odd (and 2 are even), then half of the possibilities will be odd: 128 odd numbers.

they are 24 you can make with the numbers 1-4 or any other 4 digits here they are123412431432142313421324213421432431241323412314312431423421341232413214423142134312432141324123* * * * *WRONG!These are permutations, not combinations. In a combination theorder of the digits does not matter so there is only one combination of 4 digits out of 4.

4! Four factorial. 4 * 3 * 2 = 24 permutations ------------------------

6! = 6*5*4*3*2*1 = 720

There are 6.

If you are talking about the combination function C(x,y) then your answer would be 210. However, try to be a little more specific. If you intend for permutations of 4 digit numbers (order matters), then you would have 10000 permutations for the digits, given that none of the digits are equal to each other.

If repetition is allowed and order is important, then you have essentially a base-4 number system, with the numbers ranging from 00004 to 33334. The quantity of permutations in this example is 44 = 256. If repetition is not allowed, but order is important, then it is 4! = 24. * * * * * The above answer is perfectly correct. But, as stated in the answer, for permutations. However, according to the mathematical definition of combinations (as opposed to permutations), the order is irrelevant to combinations. 1234 is the same as 1423 or 4213 etc. Consequently, there can be only one 4-number combination from 4 numbers

The answer is 6P5 = 6!/(6-5)! = 6*5*4*3*2*1/1 = 720

There are 5P3 = 5!/2! = 5*4*3 = 60 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.

Lots of them 64 which is 1296

There are 5*4*3 = 60 permutations.

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