How many ways are there to choose from a deck of cards?

Question

The prompt is How many ways are there to choose 7 cards from a deck of 52 cards, given that one card from each suit must be chosen?

The way I solved this was by imagining we already took 1 card from each suit, now we are left with 3 cards to be picked from 4 suits, this can be done easily using stars and bars giving us $${{4+3-1}\choose {3}} = 20 \space ways$$ The other way that I seemed to find on the internet that also makes sense goes like this.

• Choosing 4 from 1 suit (4+1+1+1) $${{4}\choose{1}}\cdot{{13}\choose{4}}\cdot{{13}\choose{1}}^3$$

• Choosing 2 from 3 suits (2+2+2+1) $${{4}\choose{1}}\cdot{{13}\choose{2}}^2\cdot{{13}\choose{1}}$$

• Choosing 3 from 1 suit and 2 from 2 other (3+2+1+1) $${{4}\choose{1}}\cdot{{13}\choose{4}}\cdot{{3}\choose{1}}\cdot{{13}\choose{2}}\cdot{{13}\choose{1}}^2$$

Which also makes sense but gives obviously a different answer than the first method, what am I missing? Which way is the right way?

Answer 1

The way I solved this was by imagining we already took 1 card from each suit, now we are left with 3 cards to be picked from 4 suits, this can be done easily using stars and bars giving us

Stars-and-bars counts ways to place identical items into distinct containers.   Thus it is inapplicable to counting the ways to select cards, which are distinguishable by rank.   It is the wrong way to go.

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The alternative method does consider the selection of rank.   This is the right thing to do.

$7$ can be partitioned into sums of four terms as $(4+1+1+1)$, $(3+2+1+1)$, and $(2+2+2+1)$.   We count ways to select suits for each amount, and ways to select those amounts of ranks in the selected suits.

$$\binom 4 {1,3}\binom {13}4^1\binom{13}1^3+\binom{4}{1,1,2}\binom{13}3^1\binom{13}2^1\binom{13}1^2+\binom 4{3,1}\binom{13}2^3\binom{13}1^1$$

Answer 2

This does not answer your question, but provides a route and is too large for a comment.

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Let $S$ denote the collection of sets of $7$ cards that contain at least one spade. Similarly for $D,H,C$.

The collection of sets of $7$ cards that contain no spades is denoted by $S^\complement$.

Then by inclusion/exclusion and symmetry:

$$|S^\complement\cup D^\complement\cup H^\complement\cup C^\complement|=\binom 41|S^\complement|-\binom42|S^\complement\cap D^\complement|+\binom43|S^\complement\cap D^\complement\cap H^\complement|$$leading to:$$|S\cap D\cap H\cap C|=\binom{52}7-\binom41\binom{39}7+\binom42\binom{26}7-\binom43\binom{13}7$$

Answer 3

So we have two ways to solve this question: $1.$Find the number of ways that we can choose 7 cards that does not include all the 4 suits and then subtract the answer from the number of all the possible ways that we can choose 7 cards. $2.$Find the number of ways that we can choose 7 cards from the deck with all the suits included. The second one is easier so: The number of the cards from each suit can be: $4/1/1/1$ , $3/2/1/1$ or $2/2/2/1$ I’m going to use the same approach as your second solution but please note that your second solution is also wrong because for example , in $2/2/2/1$ , it should be $P(13;11,2) ^ 3$ not $P(13;11,2) ^ 2$. I’m not going to give you the answer but please note that your second approach is right you just made a mistake in the solution. And the reason that your first solution is wrong is because you are just calculating the number of ways you can pick 3 cards from 6 cards which is obviously wrong since , let’s suppose we are using the $4/1/1/1$ combination(Let’s suppose we have 4 spades) , the number of ways you can pick 4 spades is alone bigger than your answer. Hope this helps!