Transcript Lec6Counting

Lecture 6
4.5
Generalized Permutations and
Combinations
Recap
r-permutations: The number of ways in which we can draw r balls
from a collection of n different balls, where the order
is important: P(n,r) = n! / (n-r)!
r-combinations: The number of ways in which we can draw r balls
from a collection of n different balls, where we do not care
about the ordering: C(n,r) = n! / r! (n-r) !
Today: we study counting problems where repetitions are allowed,
i.e. it is possible that the same ball is drawn multiple times.
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different
because
the slots
have labels
(distinguishable)
n balls
X
X
X
X
the balls are not replaced
when they have been drawn
r boxes
n ways
n-1 ways
n-r+1 ways
1
2
3
4
1
2
3
4
r-permutation without repetition (order important)
r-combination without repetition (order not important)
the same
because
the slots
have no labels
(indistinguishable)
we over-counted by r!
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n balls
r boxes
different
because
the slots
have labels
(distinguishable)
n ways
n ways
n ways
1
2
3
4
1
2
3
4
r-permutation with repetition (order important)
r-combination with repetition (order not important)
the balls are replaced
when they have been drawn. the same
because
Or there is a very large stack the slots
have no labels
of indistinguishable balls
(indistinguishable)
of each color.
this one is tricky !
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Example: We want to draw 2 pieces of fruit from a bowl that contains
2 apples, 2 pears, and 2 oranges. In how many ways can we do this such that:
1) The fruits are different and the order matters.
2) The fruits are different and the order does not matter.
3) The fruits are indistinguishable but the order matters.
4) The fruits are indistinguishable and the order does not matter.
 1) 6 * 5 = 30
 2) 6 * 5 / 2 = 15
P(6,2)
C(6,2)
 3) Now there are 3 kinds of fruit that we draw with replacement (since there are
enough of each kind to be able to pick any fruit at any draw). This is true
because drawing apple 1 is no different than drawing apple 2. It’s like there
are copies of the same apple present.
Thus: 3 * 3 = 9
( a a), (a p), (a o), (p a), (p p), (p o), (o a), (o p), (o o).
 4) Since the order doesn’t matter (a p) = (p a), idem (a o)=(o a), (p o)=(o p).
we over-counted 3 pieces: 9-3 = 6.
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How to count the latter (r-combination with repetition)?
One strategy could be to start from drawing where the order matters and
try to count the number of ways we over-counted (last example).
However, there is a much smarter way!
Map the problem to a bit-string as follows:
balls become
indistinguishable
indistinguishable slots
become become
distinguishable boxes.
0 1 0 0 0 1 1 1 1
C(n+r-1,r) bit-strings !
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n is number of distinct classes of objects in the original bag!
- r-permutation without repetition
- order matters
(r distinguishable slots)
- without replacement
(n distinguishable objects)
n! / (n-r)!
- r-permutation with repetition
- order matters
(r distinguishable slots)
- with replacement
(n distinct classes of indistinguishable
objects)
n^r
- r-combination without repetition
- order does not matter
(r indistinguishable slots)
- without replacement
(n distinguishable objects)
n! / r! (n-r)!
- r-combination with repetition
- order does not matter
(r indistinguishable slots)
- with replacement
(n distinct classes of indistinguishable
objects)
(n+r-1)! / r! (n-1)!
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Another problem:
Assume we have exactly n objects in a bag, and we are going to draw
all of them. The order in which we draw them is important, however
n1 objects are indistinguishable, n2 different objects are indistinguishable etc.
This is like: you can only replace the blue ball 2 times, the red ball 3 times etc.)
(b b r r r) is different from (b r b r r), but the
blue balls are the same and the red balls are the same.
(2 distinct classes of indistinguishable balls of size
2 and 3 and distinguishable slots).
1
2
4
5
3
Here we can first count the total number of permutations, pretending all balls are
different  n! (n = number of balls, not kinds of balls)
Now try to figure out which strings are equivalent:
(b1 b2 r1 r2 r3) = (b2 b1 r1 r2 r3). There are 2! ways to permute the blue balls.
(b1 b2 r2 r2 r3) = (b1 b2 r2 r1 r3). There are 3! ways to permute the red balls.
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Therefore the total number of ways is: 5!/2! 3!.
In general: n!/ n1! n2! ...nk! with n=n1+n2+...+nk
Alternate derivation:
-For the first kind we have n slots and n1 balls to place such that the
order is unimportant: C(n,n1)
-For the second kind there are then n-n1 slots still open
to place n2 balls: C(n-n1,n2) ... etc.
-For the last kind we have (n-n1-n2-...-n_(k-1)) slots for
the remaining nk balls: C(n-n1-n2-...-n_(k-1),nk).
Total thus: C(n,n1) C(n-n1,n2)..... = n!/ n1! n2! ...nk! (check it!)
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Yet another (?) problem:
In how many ways can we distribute n different objects into k different boxes
such that n1 objects go in box 1, n2 objects go in box 2,... nk objects go in box k.
(k distinguishable boxes with n distinguishable balls)
6
4
1
5
2
3
(b r r b b g)
Trick: the colors are now the boxes, so we assign boxes to balls instead of
balls to boxes! From the previous problem we know there are
n!/n1! n2!...nk! ways to assign the boxes with k colors to the n balls.
4.5 Balls in Boxes
Often, you can reverse the role of the balls and the boxes.
n distinct boxes, r balls
n different kinds of balls,
r red slots
slots are
indistinguisghable
C(n+r-1,k)
balls are indistinguishable
2
1
2
3
1
2
3
4
balls are distinct
n^r
slots are distict
6
1
2
3
4
5
slots are distict, all balls are drawn
n!/n1!...nr!
4
1
5
2
3
balls and boxes are distinct,
but number of balls in each box is known
4.5 Balls in boxes.
7
4
8
1
2
6
5
3
Distinct balls in identical boxes when the numbers of balls are known?
First imagine the boxes are distinct: n!/n1!n2!...nr!.
Then determine that we over-counted r! times: n!/n1!n2!...nr! r!
Two remaining cases: indistinguishable balls in indistinguishable boxes:
partitions. Hard to compute.
Distinguishable balls in indistinguishable boxes (where we don’t know the
number in each box) can be computed using recurrence relations.
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Examples:
1) How many ways are there to select five bills from a cash box containing
$1, $2, $5, $10, $20, $50 and $100 bills, such that the bills are indistinguishable
and the order in which they are selected is unimportant. (there are also at least
5 bills of each kind).
 This is like drawing colored balls with replacement. The colors correspond to
the values. Since the order doesn’t matter we have: C(7+5-1,5)=462
2) A cookie shop has 4 kinds of cookies and we want to pick 6. We don’t care
about the order and cookies from one kind are indistinguishable.
 Again, drawing colored balls with replacement: colors are kind of cookies.
C(6+4-1,6)=84.
3) How many solutions to x1+x2+x3=11 with xi nonnegative integers.
This is like throwing 11 balls in 3 boxes. The balls inside each box are
indistinguishable. C(11+3-1,11)=78.
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Examples:
1) In how many ways can we place 10 indistinguishable items in 8 distinguishable
boxes.
 This is precisely the problem we saw to solve the r-combination with repetition:
C(10+8-1,10)
2) How many different “words” can we create by reordering SUCCESS ?
Total number of permutations is 7!. However permuting the 3 S’s does not
create a new word, idem 2 C’s: 7!/3! 2!
3) In how many ways can we distribute a deck of 52 cards over 4 hands such that
each hand gets 5 cards?
 There are 4 boxes which receive 5 cards and 1 box which receives 32 cards.
Assign people to cards instead: 52! / (5!)^4 32!
38 p.342. Math teacher has 40 issues of a journal and packs them into 4 boxes,
10 issues each.
a) How many ways if the boxes are numbered?
 assign boxes to issues: 40! / (10!)^4
b) Now the boxes are indistinguishable.
 There are 4! ways to label the boxes,
once we have distributed them in unlabelled boxes.
Since the number of ways to distribute them in labeled boxes is given by a)
we get 40! / (10!)^4 4!.
10
40
38
7
4
9
8
1
2
6
5
3
28
27
17
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Book-Shelve Problems
4.5, Exercise 45
In how many ways can you put n different books on k different shelves?
(shelves can hold all books).
solution 1: Place books one by one. First book on k shelves. Second book
to the left or right of existing book, or on empty shelve (2+k-1=k+1).
Total: k(k+1)(k+2) ... (k+n-1) = (n+k-1)! / (k-1)!
solution 2: There are n! ways to put them into a sequence. Now we need to cut
the sequence into k subsequences using k-1 identical dividers. Imagine the dividers
look like identical books how many bit-strings are there with k-1 ones (dividers)
and n zeros (books): C(n+k-1,k-1). Total: C(n+k-1,k-1) n! = (n+k-1)!/(k-1)!