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Math127121 assignment 7solutions

Course

Discrete Mathematics  (Mathematics 1271)

23 Documents
Students shared 23 documents in this course
Academic year: 2021/2022
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Lakehead University

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Math 1271 Assignment 7 Solutions

  1. Compute

( 8

5

)

and

( 7

4

)

.

Solution:We have

( 8

5

)

=5!3!8! = 56 and

( 7

4

)

=4!3!7! = 35.

  1. How many ways are there for seven men and three women to stand in a lineso that no two women stand next to each other?

Solution:First we stand the seven men in a row. There are 7! ways to do this. Now we treat the men as dividers, creating eight places for women to stand between them. Next we select three of these places to be occupied by one woman. There are

( 8

3

)

ways to select these places. Finally, we assign the three women to the three places chosen, and there are 3! ways to do this. We have 7!

( 8

3

)

3! ways to arrange the men and women as required. Alternatively, we could view the choosing places for the women to be an ordered selection, so we would get 7!P(8,3) ways.

  1. A club with 30 members wants to elect a committee consisting of a president, a secretary, and three ordinary members. In how many ways can this be done?

Solution:First we choose the five members of the committee. There are

( 30

5

)

ways to do this. Then having chosen the five committee members, we choose an ordered pair to call president and secretary, so there are( P(5,2) ways to do this. We get 30 5

)

P(5,2) ways to choose such a committee. Alternatively, we could choose the president and secretary first, inP(30,2) ways, and then choose the three ordinary members, in

( 28

3

)

ways, for a total ofP(30,2)

( 28

3

)

ways to choose the committee. (You can quickly check these both give the same answer.)

  1. Find the coefficient ofa 10 b 6 in the expansion of (a+ 2b) 16.

Solution:We have from the binomial theorem (a+ 2b) 16 =

∑ 16

r=

( 16

r

)

(a)r(2b) 16 −r. The term with the powers of( aandbwe want is the one withr= 10 term is 16 10

)

(a) 10 (2b) 6 ,so the coefficient ofa 10 b 6 is

( 16

10

)

26.

  1. Prove the hexagon identity: ( n− 1 k− 1

)(

n k+ 1

)(

n+ 1 k

)

=

(

n− 1 k

)(

n k− 1

)(

n+ 1 k+ 1

)

for integerskandnwith 1≤k≤n− 1.

Solution:We have

L.H.=

(

(n−1)! (k−1)!((n−1)−(k−1))!

)(

n! (k+1)!(n−k−1)!

)(

(n+1)! k!(n−k+1)!

)

=(k−1)!k!(k+1)!((n−n1)!−kn−!(1)!(n+1)!n−k)!(n−k+1)!

andR.H.=

(

(n−1)! k!(n−k−1)!

)(

n! (k−1)!(n−k+1)!

)(

(n+1)! (k+1)!((n+1)−(k+1))!

)

=(k−1)!k!(k+1)!((n−n1)!−kn−!(1)!(n+1)!n−k)!(n−k+1)!=L.H.S required.

  1. Show that ifmandnare positive integers with 1≤m≤n,then m

(n m

)

=n

(n− 1 m− 1

)

1

2

Solution:We haveL.H=m

(

n! m!(n−m)!

)

=(m−1)!(n!n−m)!

andR.H=n

(

(n−1)! (m−1)!((n−1)−(m−1))!

)

=(m−n1)!((n−n1)!−m)!=(m−1)!(n!n−m)! =L.H,

as required.

  1. Supposenobjects are distributed betweenbboxes in such a way that no box is empty. How many objects must we choose in order to be sure we have emptied at least one of the boxes?

Solution: In order to have no box empty, we must haven≥b after taking objects away we still havebobjects left, then we could have one in each box and no empty boxes. Thus we could have selected as many asn−bobjects, and still failed to empty any of the boxes. Once we have removedn−b+1 objects, we must have an empty box.

  1. A drawer contains 8 red socks, 6 orange ones, and 12 green ones. How many socks must be selected to ensure having at least two of the same colour?

Solution: We consider the colours to be bins into which we place the selected socks. By the pigeon hole principle, as soon as we have 4 socks, we must have two of the same colour. If we only take 3, we could have one red, one orange, and one green, so 4 is the minimum number that will work.

  1. A tennis tournament has 16 players. How many ways are there to match the players in pairs for the first round? Suppose it is a hockey tournament, and it makes a difference which team is playing at home and which is playingaway. How many ways are there to pair the first round if there are 16 teams? Supposewe have a chess tournament, where it not only matters which board a pair is assignment to play on, but who gets the white pieces and who the black matters too. With 16 players, how many ways are there to pair the first round in this case?

Solution:We consider first the chess tournament. With both the board and colour of pieces distinct, there are 16 distinct positions to which we assign our 16 players, so there are 16! ways to pair the first round. With the hockey tournament, the colour of pieces for the chess tournament is analogous to which team of a givenpair gets to play at home, so this remains, but the cities are no longer in anyparticular order, so we have to divide by 8! to eliminate the order of the boards. We get 16!/8! ways to pair the first round of the hockey tournament. With the tennis tournament, there is no distinction between which side of the court the players are going to have, so we need to divide by 2! for each pair, but they still play on distinct courts, so we get 16!/ 28 ways to pair the tennis tournament. If we further consider the courts to be indistinguishable, we would have 16!/(2 8 8!) ways to pair the first round.

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Math127121 assignment 7solutions

Course: Discrete Mathematics  (Mathematics 1271)

23 Documents
Students shared 23 documents in this course

University: Lakehead University

Was this document helpful?
Math 1271 Assignment 7
Solutions
1. Compute 8
5and 7
4.
Solution: We have 8
5=8!
5!3! = 56 and 7
4=7!
4!3! = 35.
2. How many ways are there for seven men and three women to stand in a line so
that no two women stand next to each other?
Solution: First we stand the seven men in a row. There are 7! ways to do this.
Now we treat the men as dividers, creating eight places for women to stand between
them. Next we select three of these places to be occupied by one woman. There
are 8
3ways to select these places. Finally, we assign the three women to the three
places chosen, and there are 3! ways to do this. We have 7!8
33! ways to arrange
the men and women as required. Alternatively, we could view the choosing places
for the women to be an ordered selection, so we would get 7!P(8,3) ways.
3. A club with 30 members wants to elect a committee consisting of a president, a
secretary, and three ordinary members. In how many ways can this be done?
Solution: First we choose the five members of the committee. There are 30
5ways
to do this. Then having chosen the five committee members, we choose an ordered
pair to call president and secretary, so there are P(5,2) ways to do this. We get
30
5P(5,2) ways to choose such a committee. Alternatively, we could choose the
president and secretary first, in P(30,2) ways, and then choose the three ordinary
members, in 28
3ways, for a total of P(30,2)28
3ways to choose the committee.
(You can quickly check these both give the same answer.)
4. Find the coefficient of a10b6in the expansion of (a+ 2b)16.
Solution:We have from the binomial theorem (a+ 2b)16 =P16
r=0 16
r(a)r(2b)16−r.
The term with the powers of aand bwe want is the one with r= 10.This term is
16
10(a)10(2b)6,so the coefficient of a10b6is 16
1026.
5. Prove the hexagon identity:
n−1
k−1 n
k+ 1n+ 1
k=n−1
k n
k−1n+ 1
k+ 1
for integers kand nwith 1 ≤k≤n−1.
Solution: We have
L.H.S. =(n−1)!
(k−1)!((n−1)−(k−1))!  n!
(k+1)!(n−k−1)!  (n+1)!
k!(n−k+1)!
=(n−1)!n!(n+1)!
(k−1)!k!(k+1)!(n−k−1)!(n−k)!(n−k+1)!
and R.H.S. =(n−1)!
k!(n−k−1)!  n!
(k−1)!(n−k+1)!  (n+1)!
(k+1)!((n+1)−(k+1))!
=(n−1)!n!(n+1)!
(k−1)!k!(k+1)!(n−k−1)!(n−k)!(n−k+1)! =L.H.S. as required.
6. Show that if mand nare positive integers with 1 ≤m≤n, then
mn
m=nn−1
m−11

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