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These combinations (subsets) are enumerated by the 1 digits of the set of base 2 numbers counting from 0 to 2 n − 1, where each digit position is an item from the set of n. Given 3 cards numbered 1 to 3, there are 8 distinct combinations ( subsets ), including the empty set :
In the example C and C′ correspond to numbers 1001011001 2 = 601 10 and 1010001011 2 = 651 10, which again shows that C comes before C′. This number is not however the one one wants to represent the k -combination with, since many binary numbers have a number of raised bits different from k ; one wants to find the relative position of C in ...
For example the five compositions of 5 into distinct terms are: 5; 4 + 1; 3 + 2; 2 + 3; 1 + 4. Compare this with the three partitions of 5 into distinct terms: 5; 4 + 1; 3 + 2. Note that the ancient Sanskrit sages discovered many years before Fibonacci that the number of compositions of any natural number n as the sum of 1's and 2's is the nth ...
Rather, as explained under combinations, the number of n-multicombinations from a set with x elements can be seen to be the same as the number of n-combinations from a set with x + n − 1 elements. This reduces the problem to another one in the twelvefold way, and gives as result
Combinations and permutations in the mathematical sense are described in several articles. Described together, in-depth: Twelvefold way; Explained separately in a more accessible way: Combination; Permutation; For meanings outside of mathematics, please see both words’ disambiguation pages: Combination (disambiguation) Permutation ...
Frobenius coin problem with 2-pence and 5-pence coins visualised as graphs: Sloping lines denote graphs of 2x+5y=n where n is the total in pence, and x and y are the non-negative number of 2p and 5p coins, respectively. A point on a line gives a combination of 2p and 5p for its given total (green).
This is the minimum number of characters needed to encode a 32 bit number into 5 printable characters in a process similar to MIME-64 encoding, since 85 5 is only slightly bigger than 2 32. Such method is 6.7% more efficient than MIME-64 which encodes a 24 bit number into 4 printable characters.
Suppose East is known to have 7 spades from the bidding and after seeing dummy you deduce West to hold 2 spades; then if your two lines of play are to hope either for diamonds 5-3 or clubs 4-2, the a priori probabilities are 47% and 48% respectively but (,,,) % and (,,,) % so now the club line is significantly better than the diamond line.