The question is as follows: Tanya is filling out a lottery card and must choose six numbers out of 45. She has decided that 19 will be the third lowest number that she chooses. How many ways are there for her to do this?

I thought that the question could be answered in the following way: Let us assume that the first number that she chooses is 19 (since the order in which the numbers are arranged does not matter), which means that she has one possible choice for the first number. Then, let us assume that the next two numbers she picks are the two lowest. For the first choice she has 18 options and for the second she has 17. The next three numbers she chooses will yield 26, 25 and 24 options respectively (since the previous three numbers were the three lowest and the greatest of the three was 19). Therefore, the number of ways she can choose her numbers is: $$\ 18\times17\times26\times25\times24 \over 2!\times3!$$ However, this seems to be incorrect and I was wondering if anyone could point out where I went wrong.

  • 3
    $\begingroup$ Looks fine to me: I'd write it as $\binom{18}{2}\binom{26}{3}$. $\endgroup$ – Lord Shark the Unknown Apr 20 '17 at 15:40
  • $\begingroup$ Yup, all checks out. $\endgroup$ – user334732 Apr 20 '17 at 16:47
  • $\begingroup$ Okay, thank you. It must just be another mistake in the textbook. $\endgroup$ – F. Munnelly Apr 20 '17 at 23:14

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