There are 55 possible numbers for the first number, 54 for the second, and so on to 50 possibilities for the sixth number, giving 55 x 54 x ... x 50 different possible ways of selecting six numbers from fifty five different numbers.
However, as the order of selection does not matter (the combinations {1, 2, 3, 4, 5, 6} and {6, 5, 4, 3, 2, 1} are considered the same) the first number could appear in any of the six positions, the second in any of the remaining 5, and so on until there is only 1 place left for the last number, giving 6 x 5 x ... x 1 different ways each six digit combination will appear in the selections above, so the total number of ways of combining six numbers from fifty five different numbers is:
combinations = (55 x 54 x ... x 50) ÷ (6 x 5 x ... x 1)
= 28,989, 675
Five combinations are:
{1, 2, 3, 4, 5, 6},
{1, 2, 3, 4, 5, 7},
{1, 2, 3, 4, 5, 8},
{1, 2, 3, 4, 5, 9},
{1, 2, 3, 4, 5, 10}
I'll leave the remaining twenty eight million, nine hundred and eighty nine thousand, six hundred and seventy (28,989,670) combinations for you to list.
Where the order of selection matters, it is called a permutation. The formula to calculate the permutation of r items selected from a total of n items is:
nPr = n!/(n-r)!
Where the order of selection does not matter, it is called a combination. The formula to calculate the combination of r items selected from a total of n items is:
nCr = nPr ÷ r! = n!/(n-r)!r!
In both the above formulae the exclamation mark is the factorial of the preceding number which is the number multiplied by all positive numbers less than it, for example 5! is five factorial:
5! = 5 x 4 x 3 x 2 x 1
= 120
Number of combinations = 59*58*57*56*55/(5*4*3*2*1) = 5006386.
61
55
The numbers are: 55 and 55
55 can be divided by 1, 5, 11, and 55. These are the factors of 55, which are numbers that can divide evenly into 55 without leaving a remainder. Factors are pairs of numbers that multiply together to give the original number, in this case, 55.
26 = 64 combinations, including the null combination - which contains no numbers.
It is: 55C6 = 28,989,675
There are 11C2 = 11*10/(2*1) = 55 combinations.
Number of combinations = 59*58*57*56*55/(5*4*3*2*1) = 5006386.
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 59C5 = 59*58*57*56*55/(5*4*3*2*1) = 5,006,386 combinations.
Assuming you meant how many combinations can be formed by picking 8 numbers from 56 numbers, we have:(56 * 55 * 54 * 53 * 52 * 51 * 50 * 49)/8! = (7 * 11 * 3 * 53 * 13 * 51 * 25 * 7) = 1420494075 combinations. (Also equal to 57274321104000/40320)
I am not stupid enough to try and list them since there are 55*54*53*52*51*50/(6*5*4*3*2*1) = 28,989,675 combinations. nPr=permutation while nCr=combination. The question is how many combination of 6 are there in 55 numbers. So the answer should be based on the formula: nPr = n!/(n-r)! where ! is factorial and nCr = nPr/r! = n!/{(n-r)*r!} ; So using the formula should look likr this 55C6 = 55!/{(55-6)!*6!} = 55!/(49!*6!) = 28,989,675
There are 59C5 = 59*58*57*56*55/(5*4*3*2*1) = 5,006,386 combinations.
There are 56C5 = 56*55*54*53*51/(5*4*3*2*1) = 3,819,816 combinations.
There are 100 permutations that start with 77. There are 55 combination, though combination cannot be said to start with anything since combinations do not take account of the order of digits.
59! / 54! = 59 * 58 * 57 * 56 * 55 = 600766320