There are 5*4*3*2 = 120 such numbers.
106 or a million.
Just one. In a combination, the order of the digits does not matter.
This question does not make sense since a combination is independent of the order of its elements. That is, the combinations 1245 and 5142 are the same. Consequently, there can be no "last " numbers in a combination.
You have 6 options for the first digit, 5 for the next, 4 for the next, and 3 for the next. Multiply all those numbers together.
you could make a probability tree if you could be bothered
90
7
Just one. In a combination, the order of the digits does not matter.
106 or a million.
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.
This question does not make sense since a combination is independent of the order of its elements. That is, the combinations 1245 and 5142 are the same. Consequently, there can be no "last " numbers in a combination.
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.
The number of possible 3 digit combinations you can make out of 1-9 with outrepeated digits is:9C3 = 9!/(3!(9-3)!) = 84
You have 6 options for the first digit, 5 for the next, 4 for the next, and 3 for the next. Multiply all those numbers together.
you could make a probability tree if you could be bothered
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.
A digit is that which we used to count numbers.