The first six triangular numbers are 1, 3, 6, 10, 15 and 21.
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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.
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.
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
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.
No, rational numbers cannot r.
which r the first 10 amstrong numbers??
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.
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.
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.
By inverting the numbers. For example, if:2 x 3 = 6 then: 6 / 3 = 2
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.
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.
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
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
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.
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.
No, rational numbers cannot r.