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I have 4 decks, each with five unique cards. If I select 3 from the first, 4 from the second, 2 from the third and 2 from the fourth. In case order matters and I do not put cards back, in how many ways can I arrange the 11 cards if the order matters for each selected card?

I think that I use the following formula, for each deck, and just multiply them.

$\displaystyle\frac{n!}{(n-k)!}$

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    $\begingroup$ All 20 cards are different? $\endgroup$ – Masacroso Feb 5 '17 at 11:29
  • $\begingroup$ The cards are different $\endgroup$ – augusti Feb 5 '17 at 16:22
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I think it should be -

$4! × \binom 53 × \binom 54 × \binom 52 × \binom 52$

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I take it that the $11$ cards selected can be in all possible orders,
not necessarily with the cards of each deck together, then

Number of ways $=\dbinom53\dbinom54\dbinom52\dbinom52 \times \dfrac{11!}{3!4!2!2!}$

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  • $\begingroup$ We select first 3 cards from the first deck, then 4 from the second and so on. Order is important for each selected card. The question is about in how any ways the selected cards can be arranged. I think it should be so as I calculated it. $\endgroup$ – augusti Feb 5 '17 at 16:24
  • $\begingroup$ With your interpretation, your answer is correct, but the question mentions selection from each deck, and finally asks "how many ways can I arrange the 11 cards? ". All said and done, though, the question is definitely a bit ambiguous ! $\endgroup$ – true blue anil Feb 5 '17 at 17:03
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    $\begingroup$ It is, isn't it!? Thats what I also thought. Reread the problem again, I added "if the order matters for each selected card?" Is it still ambiguous? $\endgroup$ – augusti Feb 5 '17 at 17:07
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If order matters then instead of binomial coefficients we must use just falling factorials. In combinatorics these kind of counts are called variations without repetitions, there are a special kind of permutations, that is, choosing $k$ different objects from $n$ different objects then all possible orderings are

$$n^\underline k:=\prod_{j=0}^{k-1}(n-j)=n\cdot (n-1)\cdots (n-k+1)$$

hence from the first deck we have $n_1^\underline{k_1}$ possible outcomes, from the second $n_2^\underline{k_2}$, etc... (in your example $n_1=n_2=\ldots=5$, and $k_1=3$, $k_2=4$ and so on).

Then if you choose from your $m$ different decks with a prefixed order in the decks (as it seems in your example) then all possible orderings are

$$A:=\prod_{j=1}^m n_j^\underline{k_j}=n_1^\underline{k_1}\cdot n_2^\underline{k_2}\cdots n_m^\underline{k_m}$$

If you choose the order of the decks randomly then all possible orderings are $A\cdot m!$.

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