Number of 7-Card Hands With 5 Spades & 2 Hearts

[Walker14]

From a standard 52-card deck, how many 7-card hands have exactly 5 spades and 2 hearts?

 

[Jameson]

From a standard 52-card deck, how many 7-card hands have exactly 5 spades and 2 hearts?

Hi Walker14,

Welcome to MHB! :)

What have you tried so far? With problems like this it is very useful to use combinations. You might have seen it written like this:

\(\displaystyle {n \choose k}\)

 

[Prove It]

Hi Walker14,

Welcome to MHB! :)

What have you tried so far? With problems like this it is very useful to use combinations. You might have seen it written like this:

\(\displaystyle {n \choose k}\)

Hi Jameson, Walker14 is a student of mine. This problem came up during tutoring and I had a brain fart - unfortunately we couldn't figure out where to start, as there was so many variables to deal with - not only figuring out how many ways there are to get spades, but also how to arrange 5 of them, and then the same with hearts...

 

[Jameson]

Hi Jameson, Walker14 is a student of mine. This problem came up during tutoring and I had a brain fart - unfortunately we couldn't figure out where to start, as there was so many variables to deal with - not only figuring out how many ways there are to get spades, but also how to arrange 5 of them, and then the same with hearts...

Hi Prove It,

Ok I'll give you what I think the answer is and if you agree then you can walk your student through it in more detail. :)

Spoiler

\(\displaystyle {13 \choose 5}{13 \choose 2}\)

There are 13 cards of each suit and for one of them we want 5, the other we want 2. Multiply these together for a 7 card hand

I see the ILS is answering so hopefully he can confirm my thoughts.

 

[soroban]

Hello, Walker14!

It's easier than you think.

From a standard 52-card deck, how many 7-card hands
have exactly 5 spades and 2 hearts?

Choose 5 from the available 13 spades: $${13\choose5} = 1287$$ ways.

Choose 2 from the available 13 hearts: $${13\choose2} = 78$$ ways.

Answer: $$\;1287 \times 78 \:=\:100,\!386$$ hands.

 

[Opalg]

This had me in all sorts of trouble for a while, but I ended up realising that it is as simple as Jameson indicates. The number of arrangements is ${13 \choose 5}{13 \choose 2} = 100\,386$. In terms of probabilities, the chances of drawing such a hand from a 52-card pack are \(\displaystyle \frac{{13 \choose 5}{13 \choose 2}}{52\choose7} \approx 0.00075035565.\)

 

[Jameson]

This had me in all sorts of trouble for a while, but I ended up realising that it is as simple as Jameson indicates. The number of arrangements is ${13 \choose 5}{13 \choose 2} = 100\,386$. In terms of probabilities, the chances of drawing such a hand from a 52-card pack are \(\displaystyle \frac{{13 \choose 5}{13 \choose 2}}{52\choose7} \approx 0.00075035565.\)

These kinds of problems are either this easy or incredibly tedious. I pretty much always feel doubt with which ever way I go and I know that if it's not just multiplying some combinations together then I better set aside an afternoon to and grab a lot of paper.

 

[chisigma]

These kinds of problems are either this easy or incredibly tedious. I pretty much always feel doubt with which ever way I go and I know that if it's not just multiplying some combinations together then I better set aside an afternoon to and grab a lot of paper.

Of course, such problems are very tedious ... but put yourself in the shoes of, and I am one of them, every day for years had to solve in order to earn the salary! (Dull)...

First we look at the question as has been given by Walker 14 ...

From a standard 52-card deck, how many 7-card hands have exactly 5 spades and 2 hearts?...

The question is very simple and has nothing to do with probability and the answer is just as simple ...

$\displaystyle N= \binom {7}{5} = \binom {7}{2} = \frac{7 \cdot 6}{2!} = 21\ (1)$

If we want to tackle the problem in terms of probability must be taken into account then that is an extraction without replacement and therefore will be based on the hypergeometric distribution ...

$\displaystyle P = \binom {7}{5}\ \frac{13}{52}\ \frac{12}{51}\ \frac{11}{50}\ \frac{10}{49}\ \frac{9}{48}\ \frac{13}{47}\ \frac{12}{46} = .000750355646...\ (2)$

Kind regards

$\chi$ $\sigma$

 

[Prove It]

Thank you to all, it seems the wording of the problem was what threw us :)