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There is 1 combination of all ten numbers, 10 combinations of one number and of nine numbers, 45 combinations of two or eight numbers, 120 combinations of three or seven numbers, 210 combinations of four or six numbers and 252 combinations of five numbers. That is 1023 = 210 - 1 in total.
No. The number of permutations or combinations of 0 objects out of n is always 1. The number of permutations or combinations of 1 object out of n is always n. Otherwise, yes.
To calculate the number of 12-number combinations using numbers 1-36, we can use the formula for combinations: nCr = n! / r!(n-r)!, where n is the total number of items to choose from (36) and r is the number of items to choose (12). Plugging in the values, we get 36C12 = 36! / 12!(36-12)! = 36! / 12!24! = (363534*...25) / (121110...*1). This simplifies to 125,736,770 unique combinations.
There are 5,461,512 such combinations.
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The total number of element combinations depends on the number of elements that are being combined. For example, if you are combining 2 elements, there would be a total of 2 combinations (element 1 + element 2). If you are combining 3 elements, there would be a total of 6 combinations (element 1 + element 2, element 1 + element 3, element 2 + element 3). The formula to calculate the number of combinations is n(n-1)/2, where n is the number of elements being combined.
it is hard to say there are lot of combinations belive or not * * * * * If the previous answerer thinks 15 is a lot then true. 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. Not so hard to say!
To calculate the number of possible combinations from 10 items, you can use the formula for combinations, which is nCr = n! / r!(n-r)!. In this case, n is the total number of items (10) and r is the number of items you are choosing in each combination (which can range from 1 to 10). So, if you are considering all possible combinations (r=1 to 10), the total number of combinations would be 2^10, which is 1024.
To find the number of 5-digit combinations from 1 to 20, we first calculate the total number of options for each digit position. Since the range is from 1 to 20, there are 20 options for the first digit, 20 options for the second digit, and so on. Therefore, the total number of 5-digit combinations is calculated by multiplying these options together: 20 x 20 x 20 x 20 x 20 = 3,200,000 combinations.
There is 1 combination of all ten numbers, 10 combinations of one number and of nine numbers, 45 combinations of two or eight numbers, 120 combinations of three or seven numbers, 210 combinations of four or six numbers and 252 combinations of five numbers. That is 1023 = 210 - 1 in total.
To calculate the number of 4-digit combinations you can get from the numbers 1, 2, 2, and 6, we need to consider that the number 2 is repeated. Therefore, the total number of combinations is calculated using the formula for permutations of a multiset, which is 4! / (2!1!1!) = 12. So, there are 12 unique 4-digit combinations that can be formed from the numbers 1, 2, 2, and 6.
To calculate the number of 4-number combinations from 1 to 20, we can use the formula for combinations, which is nCr = n! / (r!(n-r)!), where n is the total number of items and r is the number of items to choose. In this case, n = 20 and r = 4. Plugging these values into the formula, we get 20C4 = 20! / (4!(20-4)!) = 4845. Therefore, there are 4845 different 4-number combinations possible from the numbers 1 to 20.
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.
To calculate the number of 3-digit combinations that can be made from the numbers 1-9, we can use the formula for permutations. Since repetition is allowed, we use the formula for permutations with repetition, which is n^r, where n is the total number of options (10 in this case) and r is the number of digits in each combination (3 in this case). Therefore, the total number of 3-digit combinations that can be made from the numbers 1-9 is 10^3 = 1000.
No. The number of permutations or combinations of 0 objects out of n is always 1. The number of permutations or combinations of 1 object out of n is always n. Otherwise, yes.
Well honey, if we're talking about 5-digit numbers from 1 to 60, you've got 60 options for the first digit, then 60 options for the second digit, and so on. So, you'd have 60 x 60 x 60 x 60 x 60, which is a whopping 7,776,000 possible combinations. Hope that clears things up for ya!