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The figure shows that 8 can be decomposed into 5 (the number of ways to climb 4 steps, followed by a single-step) plus 3 (the number of ways to climb 3 steps, followed by a double-step). The same reasoning is applied recursively until a single step, of which there is only one way to climb.
A standard exercise in elementary combinatorics is to calculate the number of ways of rolling any given value with a pair of ... {1,2,2,3,3,4} and {1,3,4,5,6,8}, as ...
Take each digit of the number (371) in reverse order (173), multiplying them successively by the digits 1, 3, 2, 6, 4, 5, repeating with this sequence of multipliers as long as necessary (1, 3, 2, 6, 4, 5, 1, 3, 2, 6, 4, 5, ...), and adding the products (1×1 + 7×3 + 3×2 = 1 + 21 + 6 = 28). The original number is divisible by 7 if and only if ...
The formula calculator concept can be applied to all types of calculator, including arithmetic, scientific, statistics, financial and conversion calculators. The calculation can be typed or pasted into an edit box of: A software package that runs on a computer, for example as a dialog box. An on-line formula calculator hosted on a web site.
A calculation is a deliberate mathematical process that transforms one or more inputs into one or more outputs or results.The term is used in a variety of senses, from the very definite arithmetical calculation of using an algorithm, to the vague heuristics of calculating a strategy in a competition, or calculating the chance of a successful relationship between two people.
For example: 24 x 11 = 264 because 2 + 4 = 6 and the 6 is placed in between the 2 and the 4. Second example: 87 x 11 = 957 because 8 + 7 = 15 so the 5 goes in between the 8 and the 7 and the 1 is carried to the 8.
The elements 2 and 1 + √ −3 are two maximal common divisors (that is, any common divisor which is a multiple of 2 is associated to 2, the same holds for 1 + √ −3, but they are not associated, so there is no greatest common divisor of a and b.
In this case the problem reduces to n − 2 people and n − 2 hats, because P 1 received h i ' s hat and P i received h 1 's hat, effectively putting both out of further consideration. For each of the n − 1 hats that P 1 may receive, the number of ways that P 2 , ..., P n may all receive hats is the sum of the counts for the two cases.