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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.

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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 $\pmatrix{1&0\\0&1}$. –  joriki Dec 2 '12 at 14:44

1 Answer 1

up vote 1 down vote accepted

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

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