The correct answer is 720. This is obtained by multiplying the binomial coefficients C(10,1), C(9,1) and C(8,1). However, if you would not like the first digit to be a zero, then the answer changes slightly. There are now 648 solutions. This is found by counting all 3 digit combinations in which zero is last, second, and not involved.
If the 6 digits can be repeated, there are 1296 different combinations. If you cannot repeat digits in the combination there are 360 different combinations. * * * * * No. That is the number of PERMUTATIONS, not COMBINATIONS. If you have 6 different digits, you can make only 15 4-digit combinations from them.
To find the number of three-digit combinations, we consider the digits from 000 to 999. Each digit can range from 0 to 9, giving us 10 options for each of the three digits. Therefore, the total number of three-digit combinations is (10 \times 10 \times 10 = 1,000).
The answer is 10C4 = 10!/[4!*6!] = 210
10C6 = 10*9*8*7/(4*3*2*1) = 210 combinations.
Each digit can appear in each of the 4 positions. There are 9 digits, therefore there are 9⁴ = 6561 such combinations.
Only one.
45 In combinations, the order of the digits does not matter so that 12 and 21 are considered the same.
Not repeating, it is 7*6*5*4 which is 840 ways ---- There are 7 choices for each of four digits, right? 74 = 2401
If the 6 digits can be repeated, there are 1296 different combinations. If you cannot repeat digits in the combination there are 360 different combinations. * * * * * No. That is the number of PERMUTATIONS, not COMBINATIONS. If you have 6 different digits, you can make only 15 4-digit combinations from them.
To find the number of three-digit combinations, we consider the digits from 000 to 999. Each digit can range from 0 to 9, giving us 10 options for each of the three digits. Therefore, the total number of three-digit combinations is (10 \times 10 \times 10 = 1,000).
120 combinations using each digit once per combination. There are 625 combinations if you can repeat the digits.
The answer is 10C4 = 10!/[4!*6!] = 210
9,000 - all the numbers between 1,000 and 9,999 inclusive. * * * * * NO. Those are PERMUTATIONS, not COMBINATIONS. Also, the question specified 4 digit combinations using 4 digits. The above answer uses 10 digits. If you start with 4 digits, you can make only 1 combination.
10C6 = 10*9*8*7/(4*3*2*1) = 210 combinations.
Only one: 2468. The order of the digits in a combination does not make a difference.
Each digit can appear in each of the 4 positions. There are 9 digits, therefore there are 9⁴ = 6561 such combinations.
The general formula is to square the number of digits it could be (in this case 5) by how many spaces there are. So in this case it would be 53 = 125 possible combinations.