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How many combinations of 6 numbers are there in 44 numbers?
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.
If using numbers 1 thru 49 how many 6 digit combinations can be made?
Using the formula n!/r!(n-r)! where n is the number of possible numbers and r is the number of numbers chosen, there are 13983816 combinations of six numbers between 1 and 49 inclusive.
What are the 6 combination numbers there in 55 numbers?
I am not stupid enough to try and list them since there are 55*54*53*52*51*50/(6*5*4*3*2*1) = 28,989,675 combinations. nPr=permutation while nCr=combination. The question is how many combination of 6 are there in 55 numbers. So the answer should be based on the formula: nPr = n!/(n-r)! where ! is factorial and nCr = nPr/r! = n!/{(n-r)*r!} ; So using the formula should look likr this 55C6 = 55!/{(55-6)!*6!} = 55!/(49!*6!) = 28,989,675
Can rational numbers r?
No, rational numbers cannot r.
How many 4 digit combinations are there for the numbers 0-5 without having the same numbers twice?
There are 360 possible combinations.Where n = how many numbers are available to choose from and r = the number chosen each time then:Number of permutations = n! / (n - r)!= 6! / (6 - 4)!= (1 * 2 * 3 * 4 * 5 * 6) / (6 - 4)!= 720 / 2! = 720 / (1 * 2) = 720 / 2 = 360.(Note: this formula only applies where the order of the numbers is important and repetition is NOT allowed.)
To generate first 10 Armstrong numbers?
which r the first 10 amstrong numbers??
How many numbers between 100 to 500 are divisible by 6?
There are 67 numbers between 100 and 500 divisible by 6. The first number greater than 100 divisible by 6: 100 ÷ 6 = 16 r 4 → first number divisible by 6 is 6 × 17 = 102 Last number less than 500 divisible by 6: 500 ÷ 6 = 83 r 2 → last number divisible by 6 is 6 × 83 = 498 → all multiples of 6 between 17 × 6 and 83 × 6 inclusive are the numbers between 100 and 500 that are divisible by 6. → there are 83 - 17 + 1 = 67 such numbers.
Three numbers in GP sum up to 21. Their product is 216. Find the numbers.?
Let a be the first of the three number and r the common ratio. Then a*ar*ar2 = a3r3 = 216 = 63 so that ar = 6 Also, a + ar + ar2 = 21 Substitute ar = 6 to give 6/r + 6 + 6r = 21 or 6 + 6r + 6r2 = 21r So that 6r2 - 15r + 6 = 0 or 2r2 - 5r + 2 = 0 2r2 - 4r - r + 2 = 0 or 2r*(r-2) - 1(r-2) = 0 or (2r-1)*(r-2) = 0 so r = 0.5 or r = 2 r = 0.5 gives the three numbers as being 12, 6, 3 and r = 2 gives 3, 6, 12.
How many combinations of 6 numbers are there in 44 numbers?
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.
How many 6 digit combinations are there in 20 numbers?
Oh, what a lovely question! Let's paint a happy little picture here. To find the number of 6-digit combinations using 20 numbers, we can use a simple formula: 20P6, which stands for 20 permutations taken 6 at a time. This gives us 387,600 unique combinations to explore and create beautiful patterns with. Just imagine all the possibilities waiting to be discovered!
How does each multiplication problem compare to its cooresponding division problem?
By inverting the numbers. For example, if:2 x 3 = 6 then: 6 / 3 = 2
If using numbers 1 thru 49 how many 6 digit combinations can be made?
Using the formula n!/r!(n-r)! where n is the number of possible numbers and r is the number of numbers chosen, there are 13983816 combinations of six numbers between 1 and 49 inclusive.
How many pin numbers are there in a adventure quest upgrade card?
There are 15 numbers but the first four numbers are always the same so if you are going to try to guess, you should find the first 4 numbers. first 4 r 8293
Dorothy is 6 years older than Ricardo The product of their present ages is twice what the product of their ages was 6 years ago How old is Dorothy?
d = r + 6 now you make the equation relating what happened 6 yrs ago, remembering that then they were r-6 and d-6: dr = 2(r-6)(d-6) plug the first eqn into the second one and solve: r(r+6) = 2(r-6)(r) you can cancel an r, since r isn't 0! r+6 = 2(r-6) r=18 So Ricardo is 18 and his sis is 24.
What are the 6 combination numbers there in 55 numbers?
I am not stupid enough to try and list them since there are 55*54*53*52*51*50/(6*5*4*3*2*1) = 28,989,675 combinations. nPr=permutation while nCr=combination. The question is how many combination of 6 are there in 55 numbers. So the answer should be based on the formula: nPr = n!/(n-r)! where ! is factorial and nCr = nPr/r! = n!/{(n-r)*r!} ; So using the formula should look likr this 55C6 = 55!/{(55-6)!*6!} = 55!/(49!*6!) = 28,989,675
Can rational numbers r?
No, rational numbers cannot r.
How many 6-digit numbers can be formed using the digits 23456 and 8 with repition?
Order does matter in this case so you need to use the formula for a Permutation with repetition allowed. This would be n^r where n=6 different numbers and r=6 possible places for each number to go when forming the full 6-digit number. 6^6 = 46,656 possible digits
