There are 15, and they are:1234 1235 1236 1245 1246 1256 1345 1346 1356 1456 2345 2346 2356 2456 3456.
The answer is 135. 15 times 9 equals 135 27 times 5 equals 135
7
There are 16C4 = 16*15*14*13/(4*3*2*1) = 1820 combinations.
6C4 = 6*5/(2*1) = 15
15
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.
There are 15, and they are:1234 1235 1236 1245 1246 1256 1345 1346 1356 1456 2345 2346 2356 2456 3456.
As the number has to start with 15, we have only 3 remaining digits to work with. There are 3 possible options for the first digit. Then out of each of these, 2 possible options for the second digit, and one option for the last. This means that in total there are 3x2x1 (6) possible combinations. These are: 15234 15243 15324 15342 15423 15432
There are infinite combinations that can make 3879
There are two combinations of numbers when multiplied together are equal to 15. 1 x 15 = 15 and 3 x 5 = 15.
If you use them only once each, you can make 15 combinations. 1 with all four digits, 4 with 3 digits, 6 with 2 digits, and 4 with 1 digit. There is also a combination containing no digits making 16 = 24 combinations from 4 elements.
There are 15C3 = 15*14*13/(3*2*1) = 455 combinations, and I am NOT even going to try listing them.
The answer is 135. 15 times 9 equals 135 27 times 5 equals 135
It is: 6C4 = 15-------------I disagree with the first answer.Each of the four digits can be chosen in 6 ways: 6 x 6 x 6 x 6 = 64 = 1296
Six fifteens and a double pair royal equals 24.
10 000 * * * * * NO! That is the number of PERMUTATIONS, not COMBINATIONS. In a combination, the order does not matter so that 1234 is the same as 1432 or 3412 etc. Assuming the 4 numbers are different, the correct answer is 15 comprising 4 1-digit combinations, 6 2-digit combinations, 4 3-digit combinations and 1 4-digit combination. Another way to look at it is that the first number can be in a combination or not. With each of these possibilities, the second can be in or out - giving 2*2 = 4 ways so far. With each of these there are two options for the third giving 2*2*2 = 8 combinations so far and then the last number makes it 2*2*2*2 = 16. But one of these combinations contains none of the numbers - each one is not in. Leaving that one out gives the answer 15. In general, the number of combinations of any size, from n distinct objects is 2n and if you exclude the null combination, it is 2n - 1.