There is only 1 combination of the numbers 123. (order does not matter)There are 6 permutations of the numbers 123. (order does matter)1 2 31 3 22 1 32 3 13 1 23 2 1
Find three consecutive positive even integers whose sum is 123 , Answer
The order of the digits in a combination does not matter. So 123 is the same as 132 or 312 etc. There are 10 combinations using just one of the digits (3 times). There are 90 combinations using 2 digits (1 once and 1 twice). There are 120 combinations using three different digit. 220 in all.
6 - 123, 132, 213, 231, 312, and 321.
To calculate the number of combinations with three numbers, you would use the formula for combinations, which is nCr = n! / r!(n-r)!. In this case, n is the total number of numbers you have to choose from, and r is the number of numbers you are choosing. So, if you have three numbers to choose from, there would be 3C3 = 3! / 3!(3-3)! = 6 / (6*0!) = 6 / 6 = 1 combination.
Just one. In combinatorials, 123 is the same as 312 and so on.
6 for example, 123 123 132 321 213 231 312 * * * * * WRONG. Those are permutations, not combinations. There are 8 possible combinations: 123 12 13 23 1 2 3 and finally, the null combination.
6
Just 4: 123, 124, 134 and 234. The order of the numbers does not matter with combinations. If it does, then they are permutations, not combinations.
Assuming the digits are not repeated, there are four combinations:123, 124, 134 and 234.
Six combinations: 123, 132, 213, 231, 312, 321
There is only 1 combination of the numbers 123. (order does not matter)There are 6 permutations of the numbers 123. (order does matter)1 2 31 3 22 1 32 3 13 1 23 2 1
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).
well, 123 213 312 321 132 231 that's all
123 132 213 231 312 321
To calculate the number of combinations with three numbers, you would use the formula for combinations, which is nCr = n! / r!(n-r)!. In this case, n is the total number of numbers you have to choose from, and r is the number of numbers you are choosing. So, if you have three numbers to choose from, there would be 3C3 = 3! / 3!(3-3)! = 6 / (6*0!) = 6 / 6 = 1 combination.
Find three consecutive positive even integers whose sum is 123 , Answer