0
What else can I help you with?
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.
How many 6-number combinations are there for numbers 1-39?
To calculate the number of 6-number combinations from a pool of numbers 1-39, we can use the combination formula, which is nCr = n! / (r!(n-r)!). In this case, n = 39 (total numbers) and r = 6 (numbers chosen). Plugging these values into the formula, we get 39C6 = 39! / (6!(39-6)!) = 3,262,623 unique combinations.
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
How many combinations are there in a set of 4 numbers?
In a set of 4 numbers, the number of combinations depends on how many numbers you want to choose from that set. If you want to choose all 4 numbers, there is only 1 combination. If you choose 2 numbers from the set, the number of combinations is calculated using the formula ( \binom{n}{r} = \frac{n!}{r!(n-r)!} ), which in this case would be ( \binom{4}{2} = 6 ). For different values of r (choosing 1, 2, or 3 numbers), the combinations would be 4, 6, and 4 respectively.
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.
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.
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 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
How many 6-number combinations are there for numbers 1-39?
To calculate the number of 6-number combinations from a pool of numbers 1-39, we can use the combination formula, which is nCr = n! / (r!(n-r)!). In this case, n = 39 (total numbers) and r = 6 (numbers chosen). Plugging these values into the formula, we get 39C6 = 39! / (6!(39-6)!) = 3,262,623 unique combinations.
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
How many combinations of 6 numbers if you have 7 numbers?
Well, honey, if you have 7 numbers and you're picking 6 of them, there are 7 ways to pick the first number, 6 ways to pick the second number, 5 ways for the third number, and so on until you've picked all 6 numbers. So, the number of combinations would be 7 x 6 x 5 x 4 x 3 x 2, which is 5040. It's like picking out outfits for the week, just with numbers instead of clothes.
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.
Can rational numbers r?
No, rational numbers cannot r.
