Combinatorics with $n$ distinct cards and $k$ people
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Essentially, given $n$ distinct cards, how many hands can I deal to $k$ people such that everyone will have at least one card? I’ve done some searching and found this related to the Sterling numbers of the second kind. However, I don’t understand how these formulas were derived, specifically how does the formula account for scenarios where someone doesn’t have a card?
As an example, say I have 17 cards and 3 people. How many ways can I deal hands to everyone such that each person has at least one card?
Relevant Information:
combinatorics
asked Jan 20 at 0:53
darylnakdarylnak
169111
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add a comment |
$begingroup$
Essentially, given $n$ distinct cards, how many hands can I deal to $k$ people such that everyone will have at least one card? I’ve done some searching and found this related to the Sterling numbers of the second kind. However, I don’t understand how these formulas were derived, specifically how does the formula account for scenarios where someone doesn’t have a card?
As an example, say I have 17 cards and 3 people. How many ways can I deal hands to everyone such that each person has at least one card?
Relevant Information:
combinatorics
asked Jan 20 at 0:53
darylnakdarylnak
169111
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You already have the formula. Why don't you just apply it?
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– Levent
Jan 20 at 1:05
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Stars and Bars provides a nice way to think about this sort of counting.
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– lulu
Jan 20 at 1:05
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@levent because I don’t understand it. I’d like to be able to explain it to someone who does not know.
$endgroup$
– darylnak
Jan 20 at 1:48
add a comment |
$begingroup$
Essentially, given $n$ distinct cards, how many hands can I deal to $k$ people such that everyone will have at least one card? I’ve done some searching and found this related to the Sterling numbers of the second kind. However, I don’t understand how these formulas were derived, specifically how does the formula account for scenarios where someone doesn’t have a card?
As an example, say I have 17 cards and 3 people. How many ways can I deal hands to everyone such that each person has at least one card?
Relevant Information:
combinatorics
asked Jan 20 at 0:53
darylnakdarylnak
169111
$endgroup$
Essentially, given $n$ distinct cards, how many hands can I deal to $k$ people such that everyone will have at least one card? I’ve done some searching and found this related to the Sterling numbers of the second kind. However, I don’t understand how these formulas were derived, specifically how does the formula account for scenarios where someone doesn’t have a card?
As an example, say I have 17 cards and 3 people. How many ways can I deal hands to everyone such that each person has at least one card?
Relevant Information:
combinatorics
combinatorics
asked Jan 20 at 0:53
darylnakdarylnak
169111
asked Jan 20 at 0:53
darylnakdarylnak
169111
asked Jan 20 at 0:53
darylnakdarylnak
169111
asked Jan 20 at 0:53
darylnakdarylnak
169111
asked Jan 20 at 0:53
darylnakdarylnak
169111
169111
$begingroup$
You already have the formula. Why don't you just apply it?
$endgroup$
– Levent
Jan 20 at 1:05
$begingroup$
Stars and Bars provides a nice way to think about this sort of counting.
$endgroup$
– lulu
Jan 20 at 1:05
$begingroup$
@levent because I don’t understand it. I’d like to be able to explain it to someone who does not know.
$endgroup$
– darylnak
Jan 20 at 1:48
add a comment |
$begingroup$
You already have the formula. Why don't you just apply it?
$endgroup$
– Levent
Jan 20 at 1:05
$begingroup$
Stars and Bars provides a nice way to think about this sort of counting.
$endgroup$
– lulu
Jan 20 at 1:05
$begingroup$
@levent because I don’t understand it. I’d like to be able to explain it to someone who does not know.
$endgroup$
– darylnak
Jan 20 at 1:48
$begingroup$
You already have the formula. Why don't you just apply it?
$endgroup$
– Levent
Jan 20 at 1:05
$begingroup$
You already have the formula. Why don't you just apply it?
$endgroup$
– Levent
Jan 20 at 1:05
$begingroup$
Stars and Bars provides a nice way to think about this sort of counting.
$endgroup$
– lulu
Jan 20 at 1:05
$begingroup$
Stars and Bars provides a nice way to think about this sort of counting.
$endgroup$
– lulu
Jan 20 at 1:05
$begingroup$
@levent because I don’t understand it. I’d like to be able to explain it to someone who does not know.
$endgroup$
– darylnak
Jan 20 at 1:48
$begingroup$
@levent because I don’t understand it. I’d like to be able to explain it to someone who does not know.
$endgroup$
– darylnak
Jan 20 at 1:48
add a comment |
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$begingroup$
You already have the formula. Why don't you just apply it?
$endgroup$
– Levent
Jan 20 at 1:05
$begingroup$
Stars and Bars provides a nice way to think about this sort of counting.
$endgroup$
– lulu
Jan 20 at 1:05
$begingroup$
@levent because I don’t understand it. I’d like to be able to explain it to someone who does not know.
$endgroup$
– darylnak
Jan 20 at 1:48