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This sounds like a Lotto-Ball type question with no repetition and order does not matter. For N items, taken R at a time, C = N! / (R!*(N-R)!) = (55*54*53*52*51*50)/(6*5*4*3*2*1) = 28,989,675 {The rest of the terms in the factorials cancel in numerator/denominator} See the related MathsIsFun link.
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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 6 number combinations in 55 numbers?
It is: 55C6 = 28,989,675
How many combination of 6 numbers are there in 42 numbers?
262
How many 6 combination numbers are there in 42 numbers what are the combinations?
677
How many whole numbers between 49 to 56?
50. 51. 52. 53. 54. 55. That is 6 numbers.
