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Tuesday 28 June 2011

Mth202 Assignment No. 5 Solution

Mth202 Assignment No. 5 Solution

Assignment#5 Of MTH202 (Spring 2011)

Maximum Marks: 15
Due Date: June 28, 2011

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Question 1; Mark: 5
A company manager wants to hire 4 people from a group of 10 applicants, 4 men and 6 women, all of whom are equally qualified to fill the position. If he selects the four at random, what is the probability that
i. All four will be women
ii. At least two will be men?

Question 2; Marks: 5
How many ways are there to select a committee to develop a discrete mathematics course at a university if the committee is to consist of four faculty members from the mathematics department and three from the computer science department, if there are 10 faculty members of the mathematics department and 12 of the computer science department?
Question 3; Marks: 5
Suppose that there are 1807 students at university. Of these, 453 are taking a course in computer science, 567 are taking a course in mathematics, and 299 are taking a courses in both computer science and mathematics .How many are not taking a course in either computer science or in mathematics?

SOLUTION:

A company manager wants to hire 4 people from a group of 10 applicants, 4 men and 6 women, all of whom are equally qualified to fill the position. If he selects the four at random, what is the probability that
  1. All four will be women
  2. At least two will be men?
C(10,4)
=10!/4!(10-4)!
=10*9*8*7*6*! / 4*3*2*1*(6!)
=5040*6*5*4*3*2*1 / 24*6*5*4*3*2*1
=5040*720 / 24*720 =362800 / 17280
=210


C(6,4)
=6!/4!(6-4)!
=6*5*4*3*2*1 / 4*3*(2!)
=720 / 24(2*1) =720/48
=15

C(8,2)
=8! / 2!(8-2)! =8*7*6*5*4*3*2*1 / 2 *1(6!)
=40320 / 2(720)
=28

Question 2;                                                                                                           Marks: 5
How many ways are there to select a committee to develop a discrete mathematics course at a university if the committee is to consist of four faculty members from the mathematics department and three from the computer science department, if there are 10 faculty members of the mathematics department and 12 of the computer science department?

C(12,4).C (10,3)
=12! / 4! (12-4)! . 10! / 3! (10-3)!
=12*11*10*9*8! / 4*3*2*1(8!). 10*9*8*7! / 3*2*1 (7!)
=11880 / 24 .720/6
=495.120
=59400

Question 3;                                                                                                           Marks: 5
Suppose that there are 1807 students at university. Of these, 453 are taking a course in computer science, 567 are taking a course in mathematics, and 299 are taking a courses in both computer science and mathematics .How many are not taking a course in either computer science or in mathematics?

N(U)=1807
N(C)=453
N(m)=567
N(C ∩ m)=299
N(cUm)=?
N(cUm)=n(C)+n (m) –n (C∩m)
=453+567-299
=721
N(cUm)-n(U) –n (cUm)
=1807-721
=1088

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