0
Add your answer:
Earn +20 pts
How many 3 number combinations can be made with 8 numbers?
Assuming that the eight numbers are all different and that none of them are zero, the number of combinations is 8C3 = 8!/(3!*5!) where n! = 1*2*3*...*n the number is 8*7*6 / (3*2*1) = 56
How many combinations does 3 tick-boxes have?
There are only four combinations but there are 8 permutations.
How many numbers can you make with 3 7 8 9?
14 * * * * * Wrong! There are 15. 4 combinations of 1 number, 6 combinations of 2 number, 4 combinations of 3 numbers, and 1 combination of 4 numbers.
How many 3 number combinations are there from 0-9 and the numbers cannot repeat?
There are 10 choices for the first number, 9 for the second and 8 for the third. 10*9*8=720 possible combinations.
How many three digit combinations 000 999 are there without repeating the same number twice?
There are 10C3 = 10*9*8/(3*2*1) = 120 combinations.
How many 3 number combinations can be made with 8 numbers?
Assuming that the eight numbers are all different and that none of them are zero, the number of combinations is 8C3 = 8!/(3!*5!) where n! = 1*2*3*...*n the number is 8*7*6 / (3*2*1) = 56
How many combinations can you make with the numbers 2 3 6 7 and 8?
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.
How many 4 number combinations are there in the number 0 to10?
The number of combinations is 10C4 = 10*9*8*7/(4*3*2*1) = 210
How many combinations does 3 tick-boxes have?
There are only four combinations but there are 8 permutations.
How many numbers can you make with 3 7 8 9?
14 * * * * * Wrong! There are 15. 4 combinations of 1 number, 6 combinations of 2 number, 4 combinations of 3 numbers, and 1 combination of 4 numbers.
How many 3 number combinations are there from 0-9 and the numbers cannot repeat?
There are 10 choices for the first number, 9 for the second and 8 for the third. 10*9*8=720 possible combinations.
What are the possible combinations of numbers 0 through 9 if any numbers cannot be repeated?
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.
How many heptagons can be drawn by joining the verticles of a polygon with 10 sides?
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.
How many three digit combinations 000 999 are there without repeating the same number twice?
There are 10C3 = 10*9*8/(3*2*1) = 120 combinations.
How many combinations can 8 and 3 get to 140?
Infinitely many. Only 6 with both, the number of 8s and the number of 3s being positive.
How many combinations of 2 and 3 digit number using 4 and 8?
the answer is = first 2-digit number by using 48= 28,82 and in 3 digit is=282,228,822,822
How many 4 number combinations are their using numbers 0 9?
10,000 * * * * * WRONG! That is the number of permutations, NOT the number of combinations. The number of combinations denoted by nCr = n!/[r!*(n-r)!] = 10!/[4!*6!] = 10*9*8*7/(4*3*2*1) = 210
