To calculate the number of combinations of 5 numbers possible from 1 to 20, we use the formula for combinations, which is nCr = n! / (r!(n-r)!). In this case, n = 20 and r = 5. Plugging these values into the formula, we get 20! / (5!(20-5)!) = 20! / (5!15!) = (20x19x18x17x16) / (5x4x3x2x1) = 15,504 possible combinations.
Oh, what a lovely question! Let's see here... To find the number of combinations of 5 numbers from 1 to 20, you can use the combination formula which is 20 choose 5. This equals 15,504 possible combinations. Isn't that just delightful? Just imagine all the beautiful possibilities waiting to be explored!
How many combinations of five. Numbers could you get from 1 to 20
There are 167960 combinations.
There are 45 combinations.
4+4+4+4+4= 20
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.
The factor pairs of 20 are pairs of numbers that can be multiplied together to give 20. The factor pairs of 20 are (1, 20), (2, 10), and (4, 5). These pairs represent all the possible combinations of numbers that can be multiplied to equal 20.
There are 167960 combinations.
There are 125970 combinations and I am not stupid enough to try and list them!
There are 167960 9 digits combinations between numbers 1 and 20.
86,450
Twenty factorial which is denoted by "20!". 20! = 1x2x3x4x5x6x7x8x...x19x20
20 and 30, among other possible combinations.
There are 45 combinations.
20 and 30, among other possible combinations.
If you have 12 possible numbers with multiple combinations then you should start out with making all the possible combinations; you will find theyre 20. Theyre four numbers out of the twleve that can be divisible by three; 3, 6, 9, and 12. There are 7 combinations where the combinations can equal those four numbers. So the odds of getting a sum divisible by three is 7/20.
Well, honey, there are 20 numbers to choose from for the first digit, 19 for the second, and 18 for the third. So, multiply those together and you get a total of 6,840 possible 3-number combinations. Math can be a real party pooper, but hey, that's the answer for ya!
There are 23C3 = 23!/(20!*3!) = 23*22*21/(3*2*1) = 1,771 combinations.
116,280