Cs201 Combinatorics Skm Asgnmnt Wid Solns

8/10/2019 Cs201 Combinatorics Skm Asgnmnt Wid Solns

1/4

1. Let A be an nxn matrix with zero in the main diagonal and 1 elsewhere. Show that determinant of A is (n-1)*(-1)^{n-1}.

As o! "now that a determinant of an nxn matrix has n# terms. $f these terms for %et(A) how man are 1& how man are ' and how man are -1

. onsider three +erm!tations of ' to ,

' 1 / 0 2 , 1 ' 0 2 , / a' a1 a a a a/ a0 a a2 a,

3ow man +erm!tations 4an be written in the row three so thatno 4ol!mn has an n!mber re+eated

. Let f(n) is a f!n4tion defined for +ositi5e integers.

6f S78{d9n} f(d) : log n& i.e.& s!m of all f(d) where d di5ides n. then determine f().

. ; 4o!+les are to be seated aro!nd a 4ir4!lar table so that men and women sit alternatel& b!t no h!sband sho!ld sit on either side of his wife. 3ow man was 4an the be seated if ; <

/. A shi++ing 4om+an has a large n!mber of wooden 4rates (boxes) in sizes 1 to 1' !nits. ard-board boxes are +a4"ed inside the4rates

to a5oid damage. =or handling 4on5enien4e& one 4rate is +a4"ed with4ard-board boxes of the same size. A 4rate is shi++ed onl when it

is4om+letel f!ll. Ass!me that dimensions of the 4rates and boxes are

s!4h that n>m boxes of size m +ro+erl fit in a 4rate of size nwhere m di5ides n. ?hen 4ard-board boxes of size m are waiting forshi++ing& +a4"ers sele4t a 4rate of size m& or m& or m ... et4.to +a4" them& with e@!al +robabilit. =ollowing table gi5es the

n!mberof 4rates shi++ed in one wee".

Size 99 1 9 9 9 9 / 9 0 9 9 2 9 , 9 1' o!nt 99 9 9 , 9 , 9 / 9 12 9 9 1' 9 10 9

3ow man 4ard-board boxes of size are shi++ed d!ring this +eriod. Show all ste+s.

Solution of Combinatorial Problem Set1

1. Let An : '11111 be the gi5en nxn matrix. Let 7n denote itsdeterminant. 1'1111 ..... 11111'


2/4

Let n : 111111 1'1111 11'111 ... 11111'

Bhe determinant of n : 7n-1 C 7n.

Show that the re4!rren4e relation 7n : -(n-1)(7n-1 C 7n-).

. Let !s label women and men as follows. Di4" an one lo4ation wherea woman is sitting. Bhis will be referred as the first lo4ation. Startingfrom the woman at this seat& label all women 1& &...& n in the 4lo4"wise order and gi5e the womanEs label to her h!sband as well. S!++osethe man sitting

to the left of woman-1 has label-a1F the man sitting to the left ofwoman- has

label a& so on. Let A : a1 a .... an. Bhen for all G& aG sho!ld

be neithere@!al to G nor sho!ld it be e@!al to GC1. Bhen ea4h +erm!tation A

of 1&...&n is a 5alid seating arrangement for men (gi5e that women are alreadseated) if A disagrees with the following +erm!tations at all +ositions.

1 : 1 ...........n-1 n : ..........n 1

So the +roblem is to find the n!mber of +erm!tations that disagreewith 1 and at all +ositions. ?e !se in4l!sion-ex4l!sion te4hni@!e to find then!mber. Let !s

define D1 to be the +erm!tation in whi4h first lo4ation has either1 or & ... Di is the set of +erm!tations in whi4h i-th lo4ation has either ior iC1& so on. 6n other words Di is the set of +erm!tations where i-th +ositionhas the 5iolation.

;r denotes the n!mber of +erm!tations in whi4h at least r +ositionsha5e

5iolations. Bo 4om+!te ;r we +ro4eed as follows.Ass!me that we 4hose 1 in the first 4ol!mn (+osition) and now we

wantto 4hoose r-1 more 5iolating 5al!es. onsider following se@!en4e

1/....n-1 n-1 n n if we 4hoose an r-1 non-4onse4!ti5e ;!mbers from it then we wo!ldha5e 4hosen

a set of r 5iolating 5al!es. Bo see it& 4onsider the followingexam+le

................................................................................ Hxam+le 1 / 0 2 , / 0 2 , 1


3/4

Let r : /. After sele4ting 1 for the first 4ol!mn 4onsider the se@!en4e //0022,, s!++ose we 4hoose x x x x then its inter+retation it as follows +artition the se@!en4e 999/9/0909292,9,9 the sele4tion gi5es 19 9 9 9 9 9 9 ,9

...............................................................................

See that + non-4onse4!ti5e obGe4ts 4an be sele4ted from ase@!en4e of m obGe4ts in (m-+C1)(+) (i.e. 4hoose + from m-+C1) was.

So in the abo5e 4ase the n!mber of was is (n-r-1)(r-1). ;owif we did the same b 4hoosing 1 for the last 4ol!mn then we will ha5e another (n-r-1)(r-1) was. Bhis 4an be re+eated b ta"ing an " as the first 5iolation.

So total n!mber of was is n.(n-r-1)(r-1). ;ow note that a+arti4!lar

4hoi4e of 5iolation will also be 4reated as &and when started at b&4&d res+e4ti5el. So the 4hoi4e of r

5iolation will be 4om+!ted r times. Bh!s the n!mber of was is (n>r)(n-r-1)(r-1). ;ow we fill in the remaining (n-r)+la4es b (n-r) n!mbers in all +ossible was. So total n!mber of was to ha5e atleast r

5iolations is

S78{i1&i&...&ir} ;i1&i&..&ir : ;r : (n>r).(n-r-1)(r-1).(n-r)#. ;ow from in4l!sion>ex4l!sion method find the n!mber of was inwhi4h no 5iolations o44!r. =inall m!lti+l the res!lt b n#>n : (n-1)# whi4h is then!mber of was in whi4h women 4o!ld seated initiall (ignoring its 4ir4!lar+erm!tation).

. ?e !se the abo5e res!lt to sol5e this +roblem. Again we !sein4l!sion-ex4l!sion method. 6n this 4ase ;r will in4l!de +erm!tations where +5iolation are in the first half and @ in the se4ond where +C@ : r. So

;r : S78{+C@:r} I(1'>+).(1'-+-1)(+-1)C(1'>@).(1'-@-1)(@-1)J.(n-r)#

;ow find the n!mber of was with no 5iolations.

/. Let f(n) denote the n!mber of 4rates of size n shi++ed d!ringthis +eriod. Let g(m) denote the n!mber of 4rates of size m 4ontaining box of sizem shi++ed.


4/4

%!e to e@!al +robabilit& g(m) 4rates of size "*m 4ontainingboxes of size m

were shi++ed& for all ". Bherefore f(n) : S78K{m9n}g(m). =ind g()from 8obi!s in5ersion. Bhe sol!tion is floor(1'>)*g() : *g().

Documents

Cs201 Combinatorics Skm Asgnmnt Wid Solns