720 permutations (the arrangement matters)
120 combinations (the arrangement doesn't matter)
10.
There are 5C3 = 5*4/(2*1) = 10 combinations
10!/3! = 604800 different combinations.
Allowing repeats, but not leading 0s, the answer is 9,000,000,000
The number of combinations you can make with the digits 1234567890 depends on how many digits you want to use and whether repetition is allowed. If you use all 10 digits without repetition, there are 10! (10 factorial) combinations, which equals 3,628,800. If you are choosing a specific number of digits (for example, 3), the number of combinations would be calculated using permutations or combinations based on the rules you set.
You could make 10*10*10*26*26*26 combinations, or 17576000 combinations.
There are 5C3 = 10 combinations.
10.
There are: 10C3 = 120
There are 5C3 = 5*4/(2*1) = 10 combinations
You can make 5 combinations of 1 number, 10 combinations of 2 numbers, 10 combinations of 3 numbers, 5 combinations of 4 numbers, and 1 combinations of 5 number. 31 in all.
There are 10 combinations of 1 number,10*9/(2*1) = 45 combinations of 2 numbers,10*9*8/(3*2*1) = 120 combinations of 3 numbers,10*9*8*7/(4*3*2*1) = 210 combinations of 4 numbers,10*9*8*7*6/(5*4*3*2*1) = 252 combinations of 5 numbers,210 combinations of 6 numbers,120 combinations of 7 numbers,45 combinations of 8 numbers,10 combinations of 9 numbers, and1 combination of 10 numbers.All in all, 210 - 1 = 1023 combinations. I have neither the time nor inclination to list them all.There are 10 combinations of 1 number,10*9/(2*1) = 45 combinations of 2 numbers,10*9*8/(3*2*1) = 120 combinations of 3 numbers,10*9*8*7/(4*3*2*1) = 210 combinations of 4 numbers,10*9*8*7*6/(5*4*3*2*1) = 252 combinations of 5 numbers,210 combinations of 6 numbers,120 combinations of 7 numbers,45 combinations of 8 numbers,10 combinations of 9 numbers, and1 combination of 10 numbers.All in all, 210 - 1 = 1023 combinations. I have neither the time nor inclination to list them all.There are 10 combinations of 1 number,10*9/(2*1) = 45 combinations of 2 numbers,10*9*8/(3*2*1) = 120 combinations of 3 numbers,10*9*8*7/(4*3*2*1) = 210 combinations of 4 numbers,10*9*8*7*6/(5*4*3*2*1) = 252 combinations of 5 numbers,210 combinations of 6 numbers,120 combinations of 7 numbers,45 combinations of 8 numbers,10 combinations of 9 numbers, and1 combination of 10 numbers.All in all, 210 - 1 = 1023 combinations. I have neither the time nor inclination to list them all.There are 10 combinations of 1 number,10*9/(2*1) = 45 combinations of 2 numbers,10*9*8/(3*2*1) = 120 combinations of 3 numbers,10*9*8*7/(4*3*2*1) = 210 combinations of 4 numbers,10*9*8*7*6/(5*4*3*2*1) = 252 combinations of 5 numbers,210 combinations of 6 numbers,120 combinations of 7 numbers,45 combinations of 8 numbers,10 combinations of 9 numbers, and1 combination of 10 numbers.All in all, 210 - 1 = 1023 combinations. I have neither the time nor inclination to list them all.
10*10
The number of combinations of 7 numbers from 10 is 10C7 = 10*9*8/(3*2*1) = 120
10!/3! = 604800 different combinations.
The answer is the number of combinations of 7 vertices from 10.That is 10C7 = 10!/(7!*3!) = 10*9*8/(3*2*1) = 120.And, there is no such word as verticles.The answer is the number of combinations of 7 vertices from 10.That is 10C7 = 10!/(7!*3!) = 10*9*8/(3*2*1) = 120.And, there is no such word as verticles.The answer is the number of combinations of 7 vertices from 10.That is 10C7 = 10!/(7!*3!) = 10*9*8/(3*2*1) = 120.And, there is no such word as verticles.The answer is the number of combinations of 7 vertices from 10.That is 10C7 = 10!/(7!*3!) = 10*9*8/(3*2*1) = 120.And, there is no such word as verticles.
Allowing repeats, but not leading 0s, the answer is 9,000,000,000