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990, but you'll need to prove it with arithmetic and logic to get the marks in the maths homework you've obviously just posted to the internet.

Q: How many 4 number combinations where the first and last numbers are the same exist?

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If the order of the numbers are important, then this is a simple combination problem. There are 10 possible numbers to choose from for the first number. Then there are 9 options for the second number. Then there are 8 options for the third, and so on. Thus, the number of possible combinations can be calculated as 10x9x8x7x6x5. This comes out at 151,200 possible combinations.

There are 10 choices for the first number, 9 for the second and 8 for the third. 10*9*8=720 possible combinations.

Well first you will start off with 000, then go up until 999. So there would be 1000 different combinations.

The first number can be any of the ten, likewise the second and the third so 10 x 10 x 10 = 1000 combinations

100. Don't forget the 00 combination. Otherwise think of all numbers between 1 and 99 with all single digits have a zero in the first column (e.g. 1 = 01) * * * * * No, that is the number of permutations! As far as combinations are concerned, 23 is the same as 32. So, you have 10*9/2 = 45 combinations.

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If the order of the numbers are important, then this is a simple combination problem. There are 10 possible numbers to choose from for the first number. Then there are 9 options for the second number. Then there are 8 options for the third, and so on. Thus, the number of possible combinations can be calculated as 10x9x8x7x6x5. This comes out at 151,200 possible combinations.

The answer will depend on how many digits there are in each of the 30 numbers. If the 30 numbers are all 6-digit numbers then the answer is NONE! If the 30 numbers are the first 30 counting numbers then there are 126 combinations of five 1-digit numbers, 1764 combinations of three 1-digit numbers and one 2-digit number, and 1710 combinations of one 1-digit number and two 2-digit numbers. That makes a total of 3600 5-digit combinations.

There are 10 choices for the first number, 9 for the second and 8 for the third. 10*9*8=720 possible combinations.

The number 1 is the first whole number of all numbers. It cannot be reduced any lower. There are no other number combinations other that 1 in the number 1.

Well first you will start off with 000, then go up until 999. So there would be 1000 different combinations.

I'm assuming they're three unique numbers. Thus, the first can be any of three, the second either of the remaining two, and the last is the last one left. Thus: combinations = 3 * 2 * 1 = 6 Or, more generally, the combinations of n numbers in such a problem is n factorial, denoted as "n!", which is every number from 1 to that number multiplied together.

The first number can be any of the ten, likewise the second and the third so 10 x 10 x 10 = 1000 combinations

100. Don't forget the 00 combination. Otherwise think of all numbers between 1 and 99 with all single digits have a zero in the first column (e.g. 1 = 01) * * * * * No, that is the number of permutations! As far as combinations are concerned, 23 is the same as 32. So, you have 10*9/2 = 45 combinations.

The first digit can be one of five integers (1 to 5) The second - fourth digits one of six integers (0 to 5). So: number of (valid) combinations is 5*6*6*6=1080 * * * * * This gives the number of permutations, not combinations. I am working on the latter.

If there are no restrictions on duplicated numbers or other patterns of numbers then there are 10 ways of selecting the first digit and also 10 ways of selecting the second digit. The number of combinations is therefore 10 x 10 = 100.

No, it isn't. In fact, it is not known whether are odd perfect numbers exist. The first perfect numbers are 6, 28, 496, and 8128.

There are many combinations of 2 numbers that sum to 70. Any combination of numbers where the first number is x and the second number is 70-x will sum to 70. For example: 69 and 1 68 and 2 67 and 3 66 and 4...