Counting the number of ways to select r objects from a list of n distinct objects: No repetition With repetition Ordered selection n! (n− r)! nr Unordered selection ( n r ) = n! (n− r)!r! ( n+ r − 1 r...

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Answered Same DayDec 15, 2021

Answer To: Counting the number of ways to select r objects from a list of n distinct objects: No repetition...

Rajeswari answered on Dec 29 2021
158 Votes
Graph theory online exam
a) We calculate solutions for all xi>0
X1+x2+x3+x4 =20
1We have solutions as 19C3 = 969
b) We are to find integer triples such that
A3<24 and a1<=2
Hence a2 >=3, a3>=3+4 =7
i.e. a3 lies between 7 and 24, including 7 but excluding 24.
i.e. th
ere are 17 solutions for a3.
A2 is less than a3-4
i.e. a2 can take values as 3,4,…20 only. Hence there are 17 solutions
a1 is less than a2-1 hence a1 can take any value from –infinty to 2 (since negative integers also included)
Together we have infinite solutions for a1, a2 and a3 satisfying the given conditions.
Qno.2
We are given a,b to lie between -2 and +2. Also a,b are integers. Hence the possibilities are
(-2,-2) (-2,-1)(-2,0) (-2,1) (-2,2)
(-1,-2) (-1,-1) (-1,0) (-1,1) (-1,2)
….
(2,-2)….(2,2)
To belong to poset (S,row) the condition should be satisfied
We find that strictly less than sum of squares is possible only for
Either (a,b) = (c,d) for a,b in -2 to 2 or
We have a^2+b^2=4 max value and 0 min value
Thus (0,0) is related to all other members.
a) (-1,2) and (2,1)
We have a=-1, b=2, c =2, d=1. We have
Hence these two are not related in the poset.
b) Maximal elements are (-2,2) (2,2) (-2,-2) (2,-2)
c) Minimal element is (0,0)
d) S has to be partitioned.
The members of S are (0,0) and this is related to all others in the poset
Whenever a and b takes values -1 or 2 we have they do not have relation.
Hence partitions can be made as where a^2+b^2 =0 one group, 1 another, 2 another, 4 another, 5 another next is 8
Thisis because sum of squares for these can take values only as 0,1,2,4,5 or 8.
So 5 parititions are possible.
Qno.3
Given that binary sequence of length n with even number of 0’s and atleast 2 ones
We have length n binary sequence with starting number 1 can be formed in
1*2^(n-1) ways
From this if we subtract the ones which does not have any 1 or have exactly 1 one, then we get the required numberof sequences with atleast 2 ones.
No of strings which does not have 1 at all = no of sequences with full zeroes = 1 (00000..)
No of strings which has exactly 1 one = 1 can be in any one of places in the n-1 positions leaving the first one
= (n-1)C1 = n-1
Hence no of sequences that contains atleast 2 ones =
Q.no.4
We have
Suppose we have un = c^n form then
We have
Solving c=2 or c=-1
Hence solution would be
, for n =0,1,2….
We have 3 = A+B and 6 = 2A-B
Solving A =3 and B=0
Q.no.5
a)
Partitions of 5 are
5
4+1
3 + 2
2 + 1 + 2
1 + 1 + 1 + 1+1
1+2+2
1+3+1
There are seven partitions
Those greater than 2 are
6 excluding 5 on the top
b.
More than 3 parts are
2+1+2
1+1+1+1+1
1+2+2
1+3+1
c) To show that the number partitions of n that have a part greater than k is the same as the number of partitions of n into more than...
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