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DISTRIBUTION PROBLEMS

When distributing nobjects to kcontainers, the objects can be distinct (e.g. different cards) or

identical (e.g. ping-pong balls). Likewise, the containers can be distinct (e.g. labeled boxes)

or identical (e.g. urns). If objects are distinct and there can be more than one object in each

container, the order ma matter (e.g. arranging books on a shelf) or it ma not (e.g. puttingballs in a bag).

Distributing How many objects can eac container get!

nobjectsto k

containers

no

restrictions

or"er

mattersat most # at $east #

at $east #%

or"er matterse&act$y #

"istinct "istinct kn k!n "(n, k) k# n!k

i"entica$ "istinct

"istinct i"entica$ "(n, i) L(n, i) "(n, k)L(n, k) $

i"entica$ i"entica$ !(n, i) !(n, k)

Distributing nDistinct Objects to kDistinct 'ontainers a b ((( n bo& # bo&

)

((( bo& *

%ach object can go to an one of the kcontainers.

"o number of distributions $ kk... k$ kn.

If each container can get at most & object, we first choose ncontainers out of kcontainers and

then put the nobjects into these containers in n# was.

"o number of distributions $ n# $ k!n.

Distributing nDistinct Objects to kDistinct 'ontainers i+ Or"er Matters

' objects a, b can be arranged in * distinct containers in $ $ &' was.+he easiest wa to understand this formula is to see that each distribution corresponds to a

permutation of a, b, , where the ' s diide the objects among different containers/

'ontainer # 'ontainer ) 'ontainer ,

a b a b

b a b a

a b a b

b a b a

a b a b

b a b a

a b a b a b a b

a b a b

b a b a

b a b a

b a b a

"ince (k0 &) s are needed, the number of distributions $ .

If each container must receie at least & object, then this is the same as putting & object into

each container first. +his can be done in n!kwas. "o total number of distributions

$ n!knumber of was of arranging n0 kobjects among kcontainers

$ n!k

$ n!k .

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Distributing nI"entica$ Objects to kDistinct 'ontainers ((( bo& # bo&

)

((( bo& *

' identical objects can be put into * distinct containers in $ $ 1 was.

+he easiest wa to understand this formula is to see that each distribution corresponds to a

permutation of o, o, , where the ' s diide the os among different containers/'ontainer # 'ontainer ) 'ontainer ,

oo oo

o o o o

o o o o

oo oo

o o o o

oo oo

"ince (k0 &) s are needed, the number of distributions $ $ .

If each container can get at most & object, then this is the same as choosingncontainers outof kcontainers to receie & object. "o number of distributions $ .

If each container must receie at least & object, then this is the same as putting & object into

each container first.

"o number of distributions $ number of was of distributing n0 kobjects to kcontainers

$

$

$ .

Distributing nDistinct Objects to kI"entica$ 'ontainers a b ((( n (((

+he number of was of distributing ndistinct objects to knon0empt identical containers is

gien b "(n, k), the "tirling numbers of the second kind.

2or example, the 3 letters a, b, c, d can be put into ' identical containers in "(3, ') $ 4

was/

a, b, c d

a, b, d c

a, c, d b

b, c, d a

a, b c, d

a, c b, d

a, d b, c

n

k

# ) , - . / 0 1 Sum

# & &

) & & '

, & * & 5

- & 4 1 & &5

. & &5 '5 &6 & 5'

/ & *& 76 15 &5 & '6*

0 & 1* *6& *56 &36 '& & 8441 & &'4 711 &46& &656 '11 '8 & 3&36

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+his table can be generated using the recurrence relation "(n, k) $ "(n0 &, k0 &) 9 k"(n0 &,

k). %ach number is e:ual to the sum of its ;northwestern neighbour and ktimes its ;northern

neighbour, where kis the column number. +his recurrence relation can be explained as

follows/

"uppose that we know the number of was to distribute n0 & distinct objects to kidentical

containers, and the number of was to distribute n0 & distinct objects to k0 & identical

containers, and we want to know the number of was of distributing nobjects to kcontainers.

We can either/

(i) start with an of the "(n0 &, k0 &) combinations and put the nth object into a new

container (there is & wa to do this), or

(ii) start with an of the "(n0 &, k) combinations and put the nth object into one of the

non0empt container (there are kwas to do this).

+hus the total number of distributions is gien b "(n, k) $ "(n0 &, k0 &) 9 k"(n0 &, k).

If some containers can be left empt, then the number of distributions $"(n, i), the

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, & & & *

- & ' & & 5

. & ' ' & & 4

/ & * * ' & & &&

0 & * 3 * ' & & &5

1 & 3 5 5 * ' & & ''2 & 3 4 1 5 * ' & & *6

#3 & 5 8 7 4 5 * ' & & 3'

+his table can be generated using the recurrence relation !(n, k) $ !(n0 &, k0 &) 9 !(n0 k,

k). %ach number is e:ual to the sum of its ;northwestern neighbour and its neighbour k

spaces aboe, where kis the column number. +his recurrence relation can be explained as

follows/

%ach k0element partition of the integer neither contains ;& or does not contain ;&.

%achk0element partition containing ;& corresponds to a

k0 & element partition of the integern0 &. 2or example, the partitions 8 $ 1 9 & 9 & $ 5 9 ' 9 & $ 3 9 * 9 & correspond to the '0

element partitions 4 $ 1 9 & $ 5 9 ' $ 3 9 *.

2or each k0element partition not containing ;&, we can subtract & from each part of the

partition. 2or example, one of the *0element partitions of 8 not containing ;&is 3 9 ' 9 '. We

can subtract & from each part to get * 9 & 9 &, which is one of the *0element partitions of 5

(which is 8 0 *). "o eer k0element partition of nnot containing ;& corresponds to a k0

element partition of the integer n0 k.

+hus the total number of distributions is gien b !(n, k) $ !(n0 &, k0 &) 9 !(n0 k, k).

If some containers can be left empt, then the number of distributions $ !(n, i).

=ef/ >ounting ?bjects In