2
$\begingroup$

Let's assume we have a lottery with $49$ balls. We draw six of them without putting them back and sort them afterwards. E.g. one possible result is ($1$,$4$,$7$,$8$,$30$,$49$).

Obviously we have $\frac{49!}{(49-6)!6!}$ different outcomes.

How many of them contain three pairs?(e.g. 2,3,7,8,15,16)

Our approach so far:

We tried to simplify this by transforming our result $(x_1,x_2,...,x_6), 1 <= x_1 < x_2 < ... < x_6 <= 49$

to

$(x_1,x_2+1,x_3+2,...,x_6+5), 1<=x_1<=x_2<=...<=x_6<=44$.

Example: $(3,5,6,20,36,49) \Rightarrow (3,4,4,17,32,44)$.

Now we can count a pair for each $x_i = x_j$. If we draw a simple graph now and add the probabilities, we get a result that doesn't match the provided solution of $\frac{43}{45402}$.

Are we on the right track? If so, where is our error?

  • 0
    I think that if you picked$48$first, you have 45 numbers left, since all you cannot pick now is 46, 47, 48, and 49. If you picked 1 first you also have 45 choices as you only lose 3 in addition to 1, 2, and 49. **Edit**: I guess this is what you meant by "at the ends we don't lose the full 5 choices".2013-08-26

0 Answers 0