0
This is an ambiguous question. Are you asking for the number of combinations, or to actually list them? Since we haven't been given the 8 numbers, I assume the former. Are you asking for the number of combinations of 4 different numbers selected from the given 8, or are duplicates allowed?
The latter is kind of complicated. The number of permutations of 4 numbers selected from 8 given numbers (or any 8 things) is very easy: since each of the 4 positions in the sequence can be any of the 8, you get 8 raised to the 4th power, which is 4096. But to get the number of combinations, you have to eliminate the duplications, and the number of duplications for a permutation depends on the number of repetitions within the permutation itself. For example, the permutation (1, 2, 1, 2) has 2 pairs; you need to count the number of combinations of the 4 positions where the 1s can go, and the other 2 positions would have 2s. That number turns out to be 6, but if there is only 1 pair, as in (1, 1, 2, 3) you have 6 pairs of positions to put the 1s in, and 2 ways of ordering the 2 and the 3, so you get 12. I could go on and on.
If you don't allow duplications, then you have 8 x 7 x 6 x 5 possible permutations, which is 1,680. Each of these can be ordered 4 x 3 x 2 x 1 ways, which is 24. Dividing 1,680 by 24 gives 70 combinations
Still curious? Ask our experts.
Chat with our AI personalities
Rafa
Taiga
BlakeAdd your answer:
Earn +20 pts
How many combinations of 4 numbers from 8 numbers?
To calculate the number of combinations of 4 numbers from 8 numbers, you would use the combination formula, which is nCr = n! / (r!(n-r)!). In this case, n = 8 and r = 4. Plugging these values into the formula, you get 8C4 = 8! / (4!(8-4)!) = 70. Therefore, there are 70 possible combinations of 4 numbers that can be chosen from a set of 8 numbers.
How many combinations for numbers 1248?
To calculate the number of combinations for the numbers 1248, we need to consider all possible arrangements of the four digits. Since all the digits are unique, there are 4 factorial (4!) ways to arrange them. This equals 4 x 3 x 2 x 1 = 24 combinations.
How many combinations of 6 numbers are there in 10 numbers?
The answer is 10!/[6!*(10-6)!] where n! represents 1*2*3*...*n Number of combinations = 10*9*8*7*6*5*4*3*2*1/(6*5*4*3*2*1*4*3*2*1) = 10*9*8*7/(4*3*2*1) = 210
How many 4 number combinations are there using numbers 0088?
Oh honey, let me break it down for you. With the numbers 0 and 8 repeated twice, you're looking at a total of 4!/(2!2!) = 6 combinations. So, you've got 0088, 0808, 0880, 8008, 8080, and 8800. That's all she wrote, darling.
How many combinations of numbers can you make with 8 and 3?
Two . . . . . 38 and 83.
