There are 11C2 = 11*10/(2*1) = 55 combinations.
There are 56C5 = 56*55*54*53*51/(5*4*3*2*1) = 3,819,816 combinations.
55C6 = 28,989,675
55C6 = 28,989,675
There are 59C5 = 59*58*57*56*55/(5*4*3*2*1) = 5,006,386 combinations.
just intrested in the number combinations * * * * * Number of combinations = 56C6 = 56*55*54*53*52*51/(6*5*4*3*2*1) = 32,468,436
There are 11C2 = 11*10/(2*1) = 55 combinations.
There are 56C5 = 56*55*54*53*51/(5*4*3*2*1) = 3,819,816 combinations.
59! / 54! = 59 * 58 * 57 * 56 * 55 = 600766320
Number of combinations = 59*58*57*56*55/(5*4*3*2*1) = 5006386.
The factors of a number are the whole numbers that can be multiplied together to equal that number. The factors of 55 are 1, 5, 11, and 55. This is because 1 x 55 = 55 and 5 x 11 = 55. These are the only combinations of whole numbers that can multiply to give the original number, making them the factors of 55.
55C6 = 28,989,675
55C6 = 28,989,675
26 = 64 combinations, including the null combination - which contains no numbers.
combinations of 55If there is not repetition in numbers used......55 x 54 x 53 x 52 x 51 * * * * *No. That is the number of permutations. For combinations the order of the number does not matter so that 1,2,3,4,5 is the same as 1,3,2,4,5 or 1,4,2,3,5 etc.There are 5*4*3*2*1 = 120 such orderings for every set of 5 numbers. So the correct answer is55*54*53*52*51 / 120 = 3478761
61
There are 59C5 = 59*58*57*56*55/(5*4*3*2*1) = 5,006,386 combinations.