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Q: How many possible different combinations of numbers r possible in just the last 4 numbers?

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Just 1.

Just 4: 123, 124, 134 and 234. The order of the numbers does not matter with combinations. If it does, then they are permutations, not combinations.

Just one. unless you count 123456 different from 132456 then there are 46656 * * * * * But you cannot count 123456 as different from 132456 since it is NOT a different combination. And the question was about combinations.

There are 1 million possible combinations. Just think of it as a sequence of numbers, starting at 000000, 000001... all the way to 999999.

whenever you have a question like this, just multiply the two numbers together and you got your answer.

10 * * * * * That is just plain wrong! It depends on how many numbers in each combination but there are 1 combination of 4 numbers out of 4, 4 combinations of 3 numbers out of 4, 6 combinations of 2 numbers out of 4, 4 combinations of 1 number out of 4. A grand total of 15 (= 24-1) combinations.

just intrested in the number combinations * * * * * Number of combinations = 56C6 = 56*55*54*53*52*51/(6*5*4*3*2*1) = 32,468,436

There is just 1 combination of 8 numbers taken 8 at a time.

There are just 5,005.There are 10C6 = 210 combinations with 6 different numbers.There are 10C5 = 252 combinations with 5 different numbers. One of these needs to be duplicated and the one to be duplicated can be chosen in 5 ways so there are 1260 of these.There are 10C4 = 210 combinations with 4 different numbers. Either one of them appears three times or two are doubled. That can be done in 10 ways so there are 2,100 of these.There are 10C3 = 120 combinations with 3 different numbers. Either one of them appears 4 times, or one twice and another three times, or all three appear twice. That can be done in 10 ways so there are 1,200 of these.There are 10C2 = 45 combinations with 2 different numbers. Either one of them appears 5 times, or one twice and another four times, or both three times. That can be done in 5 ways so there are 225 of these.There are 10C1 = 10 combinations with just 1 number repeated six times.Add these together.

I will presume that you are using the space of integers (as there are in infinite number of real or even rational numbers between 1 and 15). There are 15 integers on the interval of [1,15] and we want to find all possible combinations of 6 numbers from this set. We use a combination, 6C15= 15! / (6! * (15 - 6)!) = 15! / (6! * 9!) If you do not have a calculator which does factorials or combinations, then you can do some cancellations to make the computation a little easier: 15! = 15 * 14 * 13 * 12 * 11 * 10 * 9! so we can cancel the 9!, which leaves us with: 15 * 14 * 13 * 12 * 11 * 10 / 6! This is still going to involve the multiplication and division of very large numbers, so I took the pansy route and just used a calculator and got: 5,005 different possible combinations.

Just use combinations formula. nCr, where n=44, r=6. Plug it into the calculator or use the formula, nCr = n!/[r!(n-r)!] And you should get 7059052 as the number of combinations.

3! or Just simply 3x2x1=6 combinations * * * * * Wrong! That is the number of permutations, not combinations. There are 23 combinations: 123, 12, 13, 23, 1, 2, 3 and the null combination.

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