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A lottery ticket consists of six numbers between $1$ and $49$. A holder of a ticket wins the jackpot if the numbers match the six randomly drawn numbers. Any ticket with at least three matching numbers from the draw also qualifies for some prize money. How many different lottery tickets are there in which at least three numbers match numbers in the draw?

Given answer: $\displaystyle{\binom 66 + \binom65\binom{43}{1} + \binom64\binom{43}{2} + \binom63\binom{43}{3}}$.

I understand the presence of summation in the answer. It's due to "at least" bit in the statement of the problem. Can't seem to make sense of how they obtained the terms in the sum.

Consider the product $\displaystyle{\binom63\binom{43}{3}}$. How do we interpret it? If $6$ in the coefficient $\displaystyle{\binom63}$ is the number of places in a ticket (where three places are fixed), then shouldn't $\displaystyle{\binom63\binom{43}{3}}$ actually be $\displaystyle{\binom63\binom{46}{3}}$ given that only three numbers are taken out of $\{1, 2, \ldots, 49\}$ to fix anywhere within the six spots on a ticket? Suppose, then, $6$ in the coefficient $\displaystyle{\binom63}$ is the number of integers from $\{1, 2, \ldots, 49\}$ in which case $\displaystyle{\binom{43}{3}}$, kind of, makes sense, but then, shouldn't we choose those six integers from $\{1, 2, \ldots, 49\}$ implying presence of something like $\displaystyle{\binom{49}{6}}$ in a product term, not to mention what happens to three of six chosen integers?

Anyway, my question is How do we interpret the terms in the sum in the given answer? Thanks.

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  • $\begingroup$ There are $6$ winning numbers and $43$ losing numbers. $\endgroup$
    – saulspatz
    Jan 19, 2021 at 15:09
  • $\begingroup$ $\binom 66 + \binom65\binom{43}{1} + \binom64\binom{43}{2} + \binom63\binom{43}{3} + \binom62\binom{43}{4} + \binom61\binom{43}{5} + \binom{43}{6} =\binom{49}{6}$ but only the first four involve at least $3$ matching the $6$ winning numbers $\endgroup$
    – Henry
    Jan 19, 2021 at 15:12
  • $\begingroup$ @saulspatz, I see. That explains everything. Thanks. $\endgroup$
    – user876100
    Jan 19, 2021 at 15:13

1 Answer 1

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To have exactly three numbers correct, you choose three of the winning numbers which you can do in $6 \choose 3$ ways, then choose three losing numbers, which you can do in $43 \choose 3$ ways. As each choice of winning numbers can go with any choice of losing numbers you multiply them, getting ${6 \choose 3}{43 \choose 3}$. The other terms work the same way. To get exactly $n$ winning numbers you have ${6 \choose n}{43 \choose 6-n}$ ways because there are $6-n$ losing numbers.

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