Problem Solving Assignments

The assignment contains six questions that require mathematical solutions, including finding the number of people who speak none of three languages, calculating permutations and combinations, and solving recurrence relations. The questions cover topics such as probability, algebra, and combinatorics.

Q4.
(a)
We denote French by F and Spanish by S and German by G. Thus the details given are
like:
We have to find now the =?
Thus, it is clear from the statement that each 6 students who speak German (G) speaks
any one of the languages from French and Spanish as well.
Hence the answer is 3
(b)
But above both are not true at the same time, thus if statement (1) is true then
And if statement (2) is true then
Similarly if
Thus, number of people who speaks none of the languages is 3.
Q2.
(a)
As there are 4 red books and 5 black books and total are 9 books so the order in which the
books are placed with no restrictions is factorial of 9 that is 9!
9! = 9 *8*7*6*5*4*3*2*1
=362,880
The answer is 362,880
(b)
All black books covered must be together that means they are considered as a bundle so
their arrangement is 1! plus, the arrangement of 4 red books is 4!..
Thus the answer is (4!+1!)*5!
=( (4*3*2*1) + (1))(5*4*3*2*1)

=25 (120)
=3,000
(c)
All black books together means a bundle in itself that is 1! And all red books must be
together that means a bundle in itself that is 1! So in total 2!.
Separately red books arrangement gives 4! And separately black books arrangement
gives 5!.
Hence we get
=2! * 4! * 5!
= 2 * 24 * 120
= 48*120
=5,760
Q3.
Let the three-political party be P, Q and R
So we have some combinations like
PQR as 99 ,100, 1
Or PQR as 100, 99, 1 respectively.
This type of possible combinations occur for the value of R from 1 to 100.
Similarly for the values of P and Q same procedure is repeated so we get total number of
seats allocated to all parties with no party with overall majority as
(N(N+1))/2 which is the sum of n natural numbers from 1 to 100
Where n =100
 (100 (100+1))/2
 (100 *101)/2
 10,100/2
 5,050
Now finally multiply this by 100 as this is also repeated for P and Q parties
So, we get the final answer as
 5,050 *100
 5,05000
Q1.
(a)
To find Multiplicative inverse of a square matrix of order n*n swap the positions of a and
d in the matrix and put negative sign in front of b and c and then divide the matrix by
determinant (ad-bc)
So here we get
Thus the multiplicative inverse for this matrix is zero.
Q5.
Constraints
a ≥ 7 can be encoded as

z7 + z8 +z9 + ……
= z7 (1+ z + z2 +…..)
= z7/(1-z)
Similarly, we get for constraints b, c, d and e as (1-z35)/(1-z), z3 (1-z39)/(1-z), 1/(1-z) and
1/(1-z) respectively.
Now we have to find the solutions for non negative integers a +b +c +d +e =79 for
constraints given above.
Denoting with zn the coefficient of zn we are looking for is as
 Z79 * z7/(1-z) * (1-z35)/(1-z) * z3(1-z39)/(1-z) * (1/(1-z))2
 Z79 * z10 * (1-z35) (1-z39)/(1-z)5
 Z89 (1-z35-z39)
 (z89 – z34 -z30)

 1088430 -73815-46376
 968239
Q6.
This question needs some intial values or conditions to solve the recurrence relation