The number of combinations is nCr = (59*58*57*56*55)/(5*4*3*2*1) = 5006386
Any 5 from 59 = 5006386
-59
It's just a combination of 59 objects chosen five at a time which means (59*58*57*56*55)/(5*4*3*2*1)=5,006,386. Order does not matter in this calculation, which is why it is a combination not a permutation.
There are exactly 11 numbers with such property:15, 26, 37, 48, 59, 40, 51, 62, 73, 84, 95
59 of them. Any 2-digit number, from 41 to 99 (inclusive), rounded to the nearest 80 would lead to 80.A:It depends on what you are rounding to. Assuming you are rounding to the nearest tens it would be the following:75, 76, 77, 78, 79, 80, 81, 82, 83, & 84
9
59! / 54! = 59 * 58 * 57 * 56 * 55 = 600766320
Number of combinations = 59*58*57*56*55/(5*4*3*2*1) = 5006386.
There are 59C5 = 59*58*57*56*55/(5*4*3*2*1) = 5,006,386 combinations.
There are 59C5 = 59*58*57*56*55/(5*4*3*2*1) = 5,006,386 combinations.
The number of combinations is 59C6 which is 59!/[6!*(59-6)!] where n! = 1*2*3*...*n So 59C6 = 59*58*57*56*55*54/(6*5*4*3*2*1) = 45,057,474
There are 60C5 = 60!/(5!(60-5)!) = 60 × 59 × 58 × 57 × 56 / (5 × 4 × 3 × 2 × 1) = 5,461,512 There are 5,461,512 possible combinations of 5 numbers from 1-60.
There are no current NBA or CBA players who wear the number 59. In fact no one has ever wore the number 59. It is not a common jersey number, as many of the jersey numbers happen to feature lower numbers.
If you count zero then you have 60 numbers to choose from for the first slot.If you don't repeat that number you only have 59 for the second slot.then, you only have 58 for the third slot.that would be 205320 ways that it could happen.
Any 5 from 59 = 5006386
1 and 59.
The answer is 59C5 = 59*58*57*56*55/(5*4*3*2*1) = 5,006,386