There are: 12345C3 = 3.134847985*1011
Just one. * * * * * Depends on how many numbers are on each ring. If there are x numbers, then the total number of combinations (actually they are permutations) is x*x*x or x3.
If the question concerned the number of combinations of three different coins, the answer is 23-1 = 7. If the coins are a,b,and c, the combinations are a, b, c, ab, ac, bc, abc. If two of the coins are the same there are only 5 combinations and if all three are the same there are 3.
There are 32C3 = 32*31*30/(3*2*1) = 4960 combinations. I do not have the inclination to list them all.
You can make six different combinations: 123 132 213 231 321 312. If you just need to know the number of combinations, you could find the answer without writing everything out by letting x represent the number of digits and calculating x*(x-1).
With repeating digits, there are 33 = 27 possible combinations.Without repeating any digits, there are 6 combinations:357375537573735753
It can be divided by 3 as in 12345/3=4115, but not 9 as in 12345/9=1371.6666666666666.
yes 12345 is a prime number
To find the second smallest number that has 12345 as factors, we need to consider the prime factorization of 12345, which is 3 x 5 x 823. The smallest number with 12345 as a factor is 12345 itself. The second smallest number would be obtained by multiplying 12345 by the next smallest prime number after 3 and 5, which is 7. Therefore, the second smallest number with 12345 as factors is 12345 x 7 = 86415.
There are 6C3 = 20 such combinations.
10.
To calculate the number of 4-number combinations using the numbers 1, 2, 3, 4, and 5 without repetition, you can use the formula for permutations. Since order matters in a combination, you would use the formula for permutations, which is nPr = n! / (n - r)!. In this case, you would have 5 choices for the first number, 4 choices for the second number, 3 choices for the third number, and 2 choices for the fourth number. Therefore, the total number of 4-number combinations would be 5P4 = 5! / (5-4)! = 5 x 4 x 3 x 2 = 120 combinations.
I am assuming you mean 3-number combinations rather than 3 digit combinations. Otherwise you have to treat 21 as a 2-digit number and equate it to 1-and-2. There are 21C3 combinations = 21*20*19/(3*2*1) = 7980 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.
12345 is divisible by 3 and the answer is 4115
12345 × 3 = 37,035
You Can Create 999 Number combinations
The 120 combinations for the digits 1, 2, 3, 4, and 5 refer to the permutations of these five distinct digits. Each arrangement of these digits counts as a unique combination, and since there are 5 digits, the total number of permutations is calculated as 5! (5 factorial), which equals 120. Examples include 12345, 12354, 12435, and so on, continuing through all possible arrangements of these five numbers.