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https://math.answers.com/math-and-arithmetic/How_many_6_combination_numbers_in_55_numbers

How many 6 combination numbers in 55 numbers? - Answers

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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How many 6 combination numbers in 55 numbers? - Answers

https://math.answers.com/math-and-arithmetic/How_many_6_combination_numbers_in_55_numbers

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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https://math.answers.com/math-and-arithmetic/How_many_6_combination_numbers_in_55_numbers

How many 6 combination numbers in 55 numbers? - Answers

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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      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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