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Q: How many 4 digit password combinations can you make with the numbers 4 5 6 9 and 0?

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There are 720 of them!

90

im not geeky enough to answer that

7

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.

Just one. In a combination, the order of the digits does not matter.

106 or a million.

A digit is that which we used to count numbers.

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.

The number of possible 3 digit combinations you can make out of 1-9 with outrepeated digits is:9C3 = 9!/(3!(9-3)!) = 84

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.

With 123 digits you can make 123 one-digit numbers.

what is the greatest 4-digit number you can make with the digit 0,1,2,3, and 4

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.

you could make a probability tree if you could be bothered

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.

Not sure what a "didget" is. It is possible to make 18 5-digit numbers.

12 numbers

You could make 10*10*10*26*26*26 combinations, or 17576000 combinations.

You can make 24 4-digit numbers. 1111 2222 3333 4444 5555 6666 1234 2345 3456 1356 6543 5432 4321 1432 ok i give up!

There are 6!/3! = 120 different 3-digit numbers that can be made from these 6 digits.

Assuming you're allowed to repeat numbers, the answer is 81. For each digit, you have three choices: 2, 5, and 6. So there are three one-digit numbers you can make, and each one digit number can be combined with any of three other digits to make a two-digit number, giving 9 (3 times 3) possibilities. Now you can take any of the nine two-digit numbers as the tens and ones places in the four-digit number and any other two-digit number as the thousands and hundreds places, which gives 81 (9 times 9) possibilities. If you can't repeat numbers, the answer is zero because you don't have enough numbers to fill all the digits.

14 * * * * * Wrong! There are 15. 4 combinations of 1 number, 6 combinations of 2 number, 4 combinations of 3 numbers, and 1 combination of 4 numbers.

Their is 25 combinations

Simple enough to solve. The answer is a power of two. Assuming you have two possible digits, say for example, 3 and 4, then you simply have to multiply it by how many numbers you want to get the total number of combinations. Each number can be 3 or 4 in this case, and you have 5 numbers. That's two to the fifth. Five combinations of any two numbers. 2x2x2x2x2. The answer is 32 combinations.