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Q: How many five number combinations can you make from 5 numbers?

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You can make 5 combinations of 1 number, 10 combinations of 2 numbers, 10 combinations of 3 numbers, 5 combinations of 4 numbers, and 1 combinations of 5 number. 31 in all.

You can make 6 combinations with 3 numbers. They are: 123 213 312 132 231 321 * * * * * NO! Those are permutations! In combitorials, the order does not matter so that the combination 123 is the same as the combination 132 etc. So all of the above comprise just 1 combination. With three numbers you can have 1 combination of three numbers (as discussed above), 3 combinations of 2 numbers (12, 13 and 23) 3 combinations of 1 number (1, 2 and 3) In all, with n numbers you can have 2n - 1 combinations. Or, if you allow the null combination (that consisting of no numbers) you have 2n combinations.

You could make 10*10*10*26*26*26 combinations, or 17576000 combinations.

As a product of its prime factors: 2*5*13 = 130

If the 1, 2, 3, 4 and 5 each appear in the numbers (no repeats, no eliminations-- so numbers like 11133 are not valid) then there are 120 combinations. If you can use any of the 5 any number of times up to a max of 5 times, (any possible way of making a 5 digit number using only 1, 2, 3, 4 and 5, so that 33455 for example is valid) then the number of combinations is 3125. The 3125 combinations include the 120 combinations above. This is a permutation of 5 objects, that can be thought of as an arrangement or a rearrangement of the five numbers. The number of permutations of this 5 numbers is equal to 5!. 5! = 5 x 4 x 3 x 2 x 1 = 120 ways. Or you can use the graphing calculator to compute 5!, such as: Press 5, MATH, with the right arrow go to PRB, press 4 (for !), ENTER.

Related questions

You can make 5 combinations of 1 number, 10 combinations of 2 numbers, 10 combinations of 3 numbers, 5 combinations of 4 numbers, and 1 combinations of 5 number. 31 in all.

14 * * * * * Wrong! There are 15. 4 combinations of 1 number, 6 combinations of 2 number, 4 combinations of 3 numbers, and 1 combination of 4 numbers.

Assuming that the six numbers are different, the answer is 15.

9000

35

The rearrangement of 5 figure numbers will be 5x4x3x2x1 which is 120 combinations, when you don't repeat a number.

None. You do not have enough numbers to make even one combination.

Simple enough to solve. The answer is a power of two. Assuming you have two possible digits, say for example, 3 and 4, then you simply have to multiply it by how many numbers you want to get the total number of combinations. Each number can be 3 or 4 in this case, and you have 5 numbers. That's two to the fifth. Five combinations of any two numbers. 2x2x2x2x2. The answer is 32 combinations.

Using the combination fuction, chose three numbers from 45 numbers. The answer is 14,190.

Assuming you are treating each number as a number and not as an individual unit, the numbers you can make from these digits are 899, 989 and 998.

You can make 6 combinations with 3 numbers. They are: 123 213 312 132 231 321 * * * * * NO! Those are permutations! In combitorials, the order does not matter so that the combination 123 is the same as the combination 132 etc. So all of the above comprise just 1 combination. With three numbers you can have 1 combination of three numbers (as discussed above), 3 combinations of 2 numbers (12, 13 and 23) 3 combinations of 1 number (1, 2 and 3) In all, with n numbers you can have 2n - 1 combinations. Or, if you allow the null combination (that consisting of no numbers) you have 2n combinations.

well 5 is an odd number there for 2 numbers can't be added to make odd numbers so 4 is the closest number to five so we use it as the middle number for five :D

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