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.
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.
If no digit can be repeated then there are 5 combinations, abcd, abce, abde, acde and bcde. If you regard abdc as different from abcd then each of the 5 basic sets could be arranged 24 ways and the total would be 120 combinations.
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.
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
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.
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
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.
If no digit can be repeated then there are 5 combinations, abcd, abce, abde, acde and bcde. If you regard abdc as different from abcd then each of the 5 basic sets could be arranged 24 ways and the total would be 120 combinations.