45*44*43*42*41/(5*4*3*2*1) = 1,221,759
There are 45 such numbers.
There are 45 combinations.
45 of them.
Well, honey, if you're looking for the number of combinations of 4 numbers out of 45, it's 1,221,759. But let's be real, you're not gonna be manually counting all those combinations, so thank goodness for math and calculators. Just plug in those numbers and let the magic happen.
First figure in combination can be any of 10, second can be any of remaining 9, third any of remaining 8 and fourth any of remaining 7 so total of possibilities is 10 x 9 x 8 x 7 ie 5040.This assumes no repeating digits, in that case possibilities would be 10 to the fourth power ie 10000.----------------------------------------------------------------------------------------------The question asks for number of combinations. The above result of 5040 is the number of 4 number permutations that can be obtain from the numbers 1 to 10.The number of different 4 number combinations that can be obtain from the numbers1 to 10 is: 10C4 = 10!/(4!∙6!) = 210 different 4 number combinations.If repetition of numbers in the 4 number combination are allowed we have to ad:with 2 numbers repeated; 3∙10C3 = 3(120) = 360with 3 numbers repeated; 2∙10C2 = 2(45) = 90with 4 numbers repeated; 10with two sets of 2 numbers repeated; 10C2 = 45This gives a total of 4 number combinations with repetition of numbers allowed of:210 + 360 + 90 + 10 + 45 = 715 different 4 number combinations.
45.
Number of combinations = 45C6 = 45!/6!(45-6)! = 8,145,060
There are 45 such numbers.
About 13million,
5 digit combinations for numbers 1 to 50= 50 C 5N!= 1x2x3x4x...x (N-1) x N= (50!)/(45!x5!) =(50x49x...x2x1)/((45x44x...x2x1)(5x4x3x2x1))=(50x49x48x47x46)/(5x4x3x2)=254251200/120 =21187602118760 combinations
There are 45 combinations.
There are 45 such numbers.
Any 6 from 45 = 814506045C6 = 8145060
45
45 of them.
There are 45 of them.
45