45*44*43*42/(4*3*2*1) = 148995
if we do not want to use the same number more than once then the answer is: 49*48*47*46*45*44 however if we can use a number more than once the solution is: 49^6
Well, isn't that a happy little question! The number that goes into 44 is 1, 2, 4, 11, 22, and 44. Each of these numbers can be multiplied by another number to give you 44. Just like painting, math is all about finding the right combinations that bring harmony and balance.
Assuming a number can appear only once, there are 47C5 = 47*46*45*44*43/(5*4*3*2*1) = 1,533,939 combinations.
45!/(45!*5!) => 45*44*43*42*41/5*4*3*2*1 => 146,611,080/120 = 1,221,759
Just use combinations formula. nCr, where n=44, r=6. Plug it into the calculator or use the formula, nCr = n!/[r!(n-r)!] And you should get 7059052 as the number of combinations.
45
26 = 64 combinations, including the null combination - which contains no numbers.
44
45*44*43*42/(4*3*2*1) = 148995
1 and 11 and 21 and 31 and 42 and 12 and 22 and 32 and 43 and 13 and 23 and 33 and 44 and 14 and 24 and 34 and 416 combinations
if we do not want to use the same number more than once then the answer is: 49*48*47*46*45*44 however if we can use a number more than once the solution is: 49^6
Well, isn't that a happy little question! The number that goes into 44 is 1, 2, 4, 11, 22, and 44. Each of these numbers can be multiplied by another number to give you 44. Just like painting, math is all about finding the right combinations that bring harmony and balance.
If repetition is allowed and order is important, then you have essentially a base-4 number system, with the numbers ranging from 00004 to 33334. The quantity of permutations in this example is 44 = 256. If repetition is not allowed, but order is important, then it is 4! = 24. * * * * * The above answer is perfectly correct. But, as stated in the answer, for permutations. However, according to the mathematical definition of combinations (as opposed to permutations), the order is irrelevant to combinations. 1234 is the same as 1423 or 4213 etc. Consequently, there can be only one 4-number combination from 4 numbers
Assuming a number can appear only once, there are 47C5 = 47*46*45*44*43/(5*4*3*2*1) = 1,533,939 combinations.
45!/(45!*5!) => 45*44*43*42*41/5*4*3*2*1 => 146,611,080/120 = 1,221,759
160 games will give 3 winning numbers everytime. * * * * * I have no idea where that answewr - to a question that wasn't even asked - came from. The correct answer is 45C3 = 45*44*43/(3*2*1) = 14,190.