CS0441 Discrete Structures
Recitation 10
Xiang Xiao
The Pigeonhole Principle
Q: There are 51 houses on a street. Each house has an address between 1000 and 1099,
inclusive. Show that at least two houses have addresses that are consecutive integers.
Answer:
For integers between 1000 and 1099, the set with the most nonconsecutive integers should contain
all even or odd integers between 1000 and 1099, inclusive. In either case, the set has 50 elements.
Therefore, the set with the most nonconsecutive integers contain 50 elements.
In this case, there are 51 houses, 50 addresses can be assigned to these houses. According to the
51 
generalized pigeonhole principle, there is at least one address assigned to at least  50
  2 houses.
Because each house has an unique house address, we can not assign the same address to two
different houses.
Therefore, even the set with the most nonconsecutive integers are not enough for us to assign the
address for the 51 houses.
As a result, some houses must have address that are consecutive integers.
Permutation and Combination
Thirteen people on a softball team show up for a game.
a)
How many ways are there to choose 10 players to take the field?
The order where these 10 players are chosen does not matter.
C (13,10) 
b)
13!
 286
10!(13  10)!
How many ways are there to assign the 10 positions by selecting from the 13 peoples?
Positions are different . It matters which person is assigned to which position.
P(13,10) 
c)
13!
 1037836800
(13  10)!
Of the 13 people show up, 3 are women. How many ways are there to choose 10 players
to take the field if at least one of them must be a women.
Three situations: one women is selected; two women are selected; and three women are selected
C (3,1)  C (10,9)  C (3, 2)  C (10,8)  C (3,3)  C(10, 7)  285
C (13,10)  C (10,10)  286  1  285
Permutation and Combination
Seven women and nine men are on the faculty in the mathematics
department at a school.
a)
How many ways are there to assign a committee of five members of the
department if at least one woman must be on the committee.
C (7,1)  C (9, 4)  C(7, 2)  C (9,3)  C (7,3)  C (9, 2)  C(7, 4)  C(9,1)  C(7,5)  4242
C (16,5)  C (9,5)  4242
b) How many ways are there to select a committee of five members of the
department if at least one woman and at least one man must be on the
committee.
C (7,1)  C (9, 4)  C(7, 2)  C (9,3)  C (7,3)  C (9, 2)  C(7, 4)  C(9,1)  4221
C (16,5)  C (9,5)  C(7,5)  4221
Permutation and Combination
• How many ways are there for 10 women and six men to stand
in a line so that no two men stand next to each other?
11!
11!10!
P(10,10)  P(11, 6)  10!

(11  6)!
5!
11!
11!10!
P(10,10)  C (11, 6)  P(6, 6)  10!
 6! 
6! (11  6)!
5!
Binomial Coefficient
•
Find the coefficient of x5y8 in (x + y)13.
According to the binomial theorem, for (x + y)n the coefficient of xn-iyi should be
Therefore, the coefficient of x5y8 is:
•
(13
8 )  1287
( nj )
10
The row of Pascal’s triangle containing the binomial co-efficient k , 0  k  10 is:
( )
1
10 45 120 210 252 210 120 45 10 1
Use Pascal’s identity to produce the row immediately following this row in Pascal’s triangle.
The row immediately follow the row is:
11
11
11
11
11
11
11
11
11
11
11
(11
0 ) (1 ) ( 2 ) ( 3 ) ( 4 ) ( 5 ) ( 6 ) ( 7 ) (8 ) (9 ) (10 ) (11 )
n 1
n
n
According to the Pascal’s identity: ( k ) = ( k 1 ) + ( k )
11
10
10
10
10
, …….
We have: (11
1 ) = ( 0 ) + (1 ) = 1 + 10 = 11 , ( 2 ) = (1 ) + ( 2 ) = 10 + 45 = 55
Therefore, the row is
1
11 55 165 330 462 462 330 165 55 11 1