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 Question from Hakan, a student: Hi im a student studying advanced mathematics and i would like to know how many possible sequences of six digits there are using the numbers from 1 to 45 (1 & 45 are included) but none of the sequences can be identical even if the numbers are in different orders. Example : 1,2,3,4,5,6 2,3,4,5,6,1 these two sequences are identical even though the digits have changed their order so therefore is incorrect. I would like to know the easiest way to write these sequences along with how many there are. This is extremely important for my year 12 exams Thank You.

Hakan,

Mathematicians call these combinations. You want the number of ways of choosing 6 things from 45 things. The expression for the number of combinations of r things from n things is

nCr = n!/[r! (n - r)!]

where n!, called "n factorial" is defined by n × (n-1) × (n-2) × ··· × 2 × 1. Thus

45C6 = 45!/[6! (45 - 6)!] = 45!/[6! × 39!] = 8,145,060

I arrived at the value of 45C6 by asking Google. I went to Google and typed 45 choose 6 into the search window. Google responded with 45 choose 6 = 8 145 060.

Penny

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