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
6 of them.