Three ways (I think).
The number of possible combinations using 4 distinct numbers depends on whether the order matters and whether repetitions are allowed. If order does not matter and repetitions are not allowed, the number of combinations of 4 numbers chosen from a larger set can be calculated using the combination formula (C(n, r) = \frac{n!}{r!(n-r)!}), where (n) is the total number of numbers available. If order matters, you would use permutations instead. Please specify if you need combinations with or without repetitions and whether order matters for a more precise answer.
If only 1 operation is happening, always go from left to right. it does matter with subtraction.
The number of different 4-letter combinations depends on whether you allow repetitions and whether the order matters. If repetitions are allowed and the order matters, there are 26^4 (or 456,976) combinations using the 26 letters of the English alphabet. If repetitions are not allowed and the order matters, there are 26 × 25 × 24 × 23 = 358,800 combinations. If order doesn't matter, the calculation changes further, but typically the first two scenarios are most common.
6 if order doesn't matter
There are: 8C6 = 8!/6!(8-6)! = (8 x 7)/(2 x 1) = 28.
communative property is when you are adding or subtracting any numbers it doesnt matter how u write them.....
The number of possible combinations using 4 distinct numbers depends on whether the order matters and whether repetitions are allowed. If order does not matter and repetitions are not allowed, the number of combinations of 4 numbers chosen from a larger set can be calculated using the combination formula (C(n, r) = \frac{n!}{r!(n-r)!}), where (n) is the total number of numbers available. If order matters, you would use permutations instead. Please specify if you need combinations with or without repetitions and whether order matters for a more precise answer.
If only 1 operation is happening, always go from left to right. it does matter with subtraction.
The number of different 4-letter combinations depends on whether you allow repetitions and whether the order matters. If repetitions are allowed and the order matters, there are 26^4 (or 456,976) combinations using the 26 letters of the English alphabet. If repetitions are not allowed and the order matters, there are 26 × 25 × 24 × 23 = 358,800 combinations. If order doesn't matter, the calculation changes further, but typically the first two scenarios are most common.
6 if order doesn't matter
There are: 8C6 = 8!/6!(8-6)! = (8 x 7)/(2 x 1) = 28.
Because if you rotate it about its center it looks different (not symmetric) no matter what the angle is.
There are infinitely many rational numbers between any two (different) numbers, no matter how close together they are.
it doesnt matter some might love it some might like it it doesnt matter
There is no commutative property in subtraction or division because the order of the numbers cannot be change. This means that when multiplying or adding it does not matter the order of the numbers because the answer comes out the same.
Because it doesn't matter what numbers you select to start the tree. If you do it correctly, the bottom branch will always have the same numbers.
it doesnt matter