Well, honey, there are 20 numbers to choose from for the first digit, 19 for the second, and 18 for the third. So, multiply those together and you get a total of 6,840 possible 3-number combinations. Math can be a real party pooper, but hey, that's the answer for ya!
The first number has 19 possibilities. The second has 18, and the third only 17. 19x18x17=5814 possible lock combinations. (If you could repeat, it would be 19x19x19=6859)
There are two ambiguities in this question. First, the numbers 0 through 9 could mean integers or real numbers. If you meant real numbers the answer is infinite, so I presume you mean integers. More to the point, it depends on whether you are only counting combinations of 4 different numbers or allowing duplications (like 4, 4, 5, 5).
there are many combinations of numbers that add up to 84. Some examples are 41+43, 35+49, 29+55, 24+61, and 18+66.
That would be: 20! / 10! = 20 * 19 * 18 * 17 * 16 * 15 * 14 * 13 * 12 * 11 = 670442572800 * * * * * No! That is the number of permutations! The number of combinations is 20C10 = 20!/(10!*10!) = 20*19*18*17*16*15*14*13*12*11/(10*9*8*7*6*5*4*3*2*1) = 184,756
121
The first number has 19 possibilities. The second has 18, and the third only 17. 19x18x17=5814 possible lock combinations. (If you could repeat, it would be 19x19x19=6859)
18 protons as its atomic number is 18
There are two ambiguities in this question. First, the numbers 0 through 9 could mean integers or real numbers. If you meant real numbers the answer is infinite, so I presume you mean integers. More to the point, it depends on whether you are only counting combinations of 4 different numbers or allowing duplications (like 4, 4, 5, 5).
The number of combinations is 20C5 = 20!/(15!*5!) = 20*19*18*17*16/(5*4*3*2*1) = 15,504
There are infinitely many numbers whose difference is 18 . Every number,say n, differs by 18 from n+ 18 and n -18.
there are many combinations of numbers that add up to 84. Some examples are 41+43, 35+49, 29+55, 24+61, and 18+66.
To calculate the number of combinations of 5 numbers possible from 1 to 20, we use the formula for combinations, which is nCr = n! / (r!(n-r)!). In this case, n = 20 and r = 5. Plugging these values into the formula, we get 20! / (5!(20-5)!) = 20! / (5!15!) = (20x19x18x17x16) / (5x4x3x2x1) = 15,504 possible combinations.
That would be: 20! / 10! = 20 * 19 * 18 * 17 * 16 * 15 * 14 * 13 * 12 * 11 = 670442572800 * * * * * No! That is the number of permutations! The number of combinations is 20C10 = 20!/(10!*10!) = 20*19*18*17*16*15*14*13*12*11/(10*9*8*7*6*5*4*3*2*1) = 184,756
233 = 18 combinations.
18. 18 x 3 = 54. However, since the question asks for even numbers, many other combinations would qualify (e.g. 40 + 8 + 6, 32 + 12 = 10).
Both.