Q: How many 4 number combinations are there in numbers 1 to 18?

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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

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).

Related questions

121

How many combinations of five. Numbers could you get from 1 to 20

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.

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.