To find all possible numbers using the digits 3, 5, 6, and 1, we can use the concept of permutations. There are 4 digits available, so there are 4 factorial (4!) ways to arrange them, which equals 24 possible combinations. This means there are 24 different numbers that can be formed using the digits 3, 5, 6, and 1.
Without specifying any limits on the number of uses of each digit and how many digits can be in the answer there are an infinite number of number that can be made using the digits {3, 5, 6, 1}.
If the number has to be 4 digits long and each digit can only be used once, there are 24 numbers:
1356, 1365, 1536, 1563, 1635, 1653,
3156, 3165, 3516, 3561, 3615, 3651,
5136, 5163, 5316, 5361, 5613, 5631,
6135, 6153, 6315, 6351, 6513, 6531.
If the number is 4 digits long and a digit can be repeated, then there are 256 numbers:
1111, 1113, 1115, 1116, 1131, 1133, 1135, 1136, 1151, 1153, 1155, 1156, 1161, 1163, 1165, 1166,
1311, 1313, 1315, 1316, 1331, 1333, 1335, 1336, 1351, 1353, 1355, 1356, 1361, 1363, 1365, 1366,
1511, 1513, 1515, 1516, 1531, 1533, 1535, 1536, 1551, 1553, 1555, 1556, 1561, 1563, 1565, 1566,
1611, 1613, 1615, 1616, 1631, 1633, 1635, 1636, 1651, 1653, 1655, 1656, 1661, 1663, 1665, 1666,
3111, 3113, 3115, 3116, 3131, 3133, 3135, 3136, 3151, 3153, 3155, 3156, 3161, 3163, 3165, 3166,
3311, 3313, 3315, 3316, 3331, 3333, 3335, 3336, 3351, 3353, 3355, 3356, 3361, 3363, 3365, 3366,
3511, 3513, 3515, 3516, 3531, 3533, 3535, 3536, 3551, 3553, 3555, 3556, 3561, 3563, 3565, 3566,
3611, 3613, 3615, 3616, 3631, 3633, 3635, 3636, 3651, 3653, 3655, 3656, 3661, 3663, 3665, 3666,
5111, 5113, 5115, 5116, 5131, 5133, 5135, 5136, 5151, 5153, 5155, 5156, 5161, 5163, 5165, 5166,
5311, 5313, 5315, 5316, 5331, 5333, 5335, 5336, 5351, 5353, 5355, 5356, 5361, 5363, 5365, 5366,
5511, 5513, 5515, 5516, 5531, 5533, 5535, 5536, 5551, 5553, 5555, 5556, 5561, 5563, 5565, 5566,
5611, 5613, 5615, 5616, 5631, 5633, 5635, 5636, 5651, 5653, 5655, 5656, 5661, 5663, 5665, 5666,
6111, 6113, 6115, 6116, 6131, 6133, 6135, 6136, 6151, 6153, 6155, 6156, 6161, 6163, 6165, 6166,
6311, 6313, 6315, 6316, 6331, 6333, 6335, 6336, 6351, 6353, 6355, 6356, 6361, 6363, 6365, 6366,
6511, 6513, 6515, 6516, 6531, 6533, 6535, 6536, 6551, 6553, 6555, 6556, 6561, 6563, 6565, 6566,
6611, 6613, 6615, 6616, 6631, 6633, 6635, 6636, 6651, 6653, 6655, 6656, 6661, 6663, 6665, 6666
Total number of possible 3-digit numbers = 9!x10!10!
7
Yes, it is. 5/1 equals 5.
Using the formula n!/r!(n-r)! where n is the number of possible numbers and r is the number of numbers chosen, there are 13983816 combinations of six numbers between 1 and 49 inclusive.
The four digits can be used to produce infinitely many different numbers if repetition is permitted. Without repetition, there are 24 possible numbers. A lot more can be produced if the numbers are combined using binary oprations, fore example, 19 * 8/4 = 19*2 = 38.
Total number of possible 3-digit numbers = 9!x10!10!
7
One possible pair of numbers is 1 and 60.
Yes, it is. 5/1 equals 5.
Using the formula n!/r!(n-r)! where n is the number of possible numbers and r is the number of numbers chosen, there are 13983816 combinations of six numbers between 1 and 49 inclusive.
Using various mathematical techniques, this is possible.
Using only sums and differences, and not necessarily all four numbers, 1, 3, 9 and 27 will make all numbers from 0 to 40.
The four digits can be used to produce infinitely many different numbers if repetition is permitted. Without repetition, there are 24 possible numbers. A lot more can be produced if the numbers are combined using binary oprations, fore example, 19 * 8/4 = 19*2 = 38.
Impossible, as there are not enough numbers to cover all squares.
Just 1.
46
21+3 or using all the numbers: 21 + (3*(2-1))