There are 18C4 = 18!/[18-4)!4!] = 18*17*16*15/(4*3*2*1) = 3060 combinations.
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 pairs of numbers that add up to 18, such as 10 and 8, 12 and 6, or 15 and 3. Additionally, you can have combinations like 9 and 9, or even include negative and decimal numbers, such as 20 and -2. Ultimately, any two numbers that, when combined, total 18 will work.
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.
121
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)
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).
There are many pairs of numbers that add up to 18, such as 10 and 8, 12 and 6, or 15 and 3. Additionally, you can have combinations like 9 and 9, or even include negative and decimal numbers, such as 20 and -2. Ultimately, any two numbers that, when combined, total 18 will work.
There are infinitely many numbers whose difference is 18 . Every number,say n, differs by 18 from n+ 18 and n -18.
The number of combinations is 20C5 = 20!/(15!*5!) = 20*19*18*17*16/(5*4*3*2*1) = 15,504
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.
The number 18 can be represented as a sum of arrays in multiple ways, depending on the constraints such as the number of elements in the array or the range of numbers allowed. For example, if considering positive integers, one can break it down into parts like (1, 17), (2, 16), and so on, including combinations like (9, 9), which ultimately leads to a combinatorial problem. In total, there are 54 different combinations of positive integers that sum to 18 when including permutations. The exact count can vary with the specific rules applied.
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