0
$\begingroup$

Good day everyone.

I'm doing a combination problem and its solution keeps eluding me.

Problem:

Of a hand of 13 in a deck of 52 how many combinations are there of none of the cards being greater than 10, assuming that ace is greater than 10.

Attempted solution:

There are sixteen cards greater than 10: ace, jack, queen and king (all houses). So of thirteen cards in a house four are too great.

\begin{equation} \left( \begin{array}{c} 13 \\ 4 \end{array} \right) \end{equation}

Possible combinations of a hand consisting of only higher than 10 cards is:

\begin{equation} \left( \begin{array}{c} 16 \\ 13 \end{array} \right) \end{equation}

Multiply them together and subtract that from

\begin{equation} \left( \begin{array}{c} 52 \\ 13 \end{array} \right) \end{equation}

to get the wrong answer. There is something missing or wrong or both and I would appreciate help.

Thanks for your time.

  • 0
    Note that you can produce binomial coefficients more easily using e.g. `\binom nk` to produce $\binom nk$. Also, general matrices in parentheses are more easily produced using e.g. `\pmatrix{1&0\\0&1}` rather than `\left(\begin{array}{c}1&0\\0&1\end{array}\right)` to produce \pmatrix{1&0\\0&1}.2012-12-02

1 Answers 1

1

You don't really need to get into the substraction here. You can count directly the "good" hands. You've already shown there are $16$ "bad" cards. That leaves $52-16=36$ good cards. You need to pick $13$ of them. You thus get $ \binom {36}{13} $ "good" hands