First you need to find you first number. As there are five numbers, the first number can be any of the five. So you have five combinations (5). Next you take your second number, as you've already used one you can only choose one of four numbers (5 x 4). You third number, you can only choose one of three numbers (5 x 4 x 3), you fourth, one of two (5 x 4 x 3 x 2) and finally you are left with one number, so you only have one choice.
So we have a combination of 5 x 4 x 3 x 2 x 1 (or 5!) to find the number of different 5 number combinations you can make with 1, 2, 3, 4 and 5.
Answer: 120.
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There are infinitely many numbers and so infinitely many possible combinations.
The number of combinations - not to be confused with the number of permutations - is 2*21 = 42.
If the numbers can be repeated and the numbers are 0-9 then there are 1000 different combinations.
If the order of the numbers are important, then this is a simple combination problem. There are 10 possible numbers to choose from for the first number. Then there are 9 options for the second number. Then there are 8 options for the third, and so on. Thus, the number of possible combinations can be calculated as 10x9x8x7x6x5. This comes out at 151,200 possible combinations.
In a 7 segment display, the symbols can be created using a selected number of segments where each segment is treated as a different element.When 1 segment is used, the possible positions are 7because it can be any of the 7 segments (7C1=7).When 2 segments are used, the number of possible combinations are 7C2=21.When 3 segments are used, the number of possible combinations are 7C3=35When 4 segments are used, the number of possible combinations are 7C4=35When 5 segments are used, the number of possible combinations are 7C5=21When 6 segments are used, the number of possible combinations are 7C6=7When 7 segments are used, the number of possible combinations are 7C7=1Adding the combinations, 7+21+35+21+7+1=127Therefore, 127 symbols can be made using a 7 segment display!