Formula of Circular Arrangement

Running head: MATHEMATICS 1
Discrete Mathematics
Name
Institution

MATHEMATICS 2
Solutions
Question 1
Let A, B, and C be sets of integers between 1 and 100000(inclusive) which are divisible by 31,
607, and 1901 respectively. As a result, we want to find;
( AUBUC )C
= |AUBUC| = |A| + |B| + |C | -( AnB ) - ( AnC ) -( AnBnC )
Where;
|A| = 100,000
31 = 3225
|B| = 100,000
607 = 164
|C | = 100,000
1901 = 52
( AnB ) = 100,000
18817 =5
( AnC ) = 100,000
58931 =1
| AUBUC| = 3225+164+52-5-1-2
= 96563
Question 2
Using the formula of circular arrangement of n-objects i.e. (n-1)!

MATHEMATICS 3
9 women will be (9-1)! = 8!, r =6, thus;
P (n,r) = n!
(n−r )! = 9 !
3 ! = 60480
=40320*60480
=2438553600 ways
Question 3
The main objective is to find the numbers of ways of different seating arrangements
=( 16
10 )
Circular arrangement for n-objects will therefore be;
(n-1)!
=(10-1)!
Remaining six people to be arranged on the six seats i.e. (6-1)!
Using the rule of product, we get;
(16
10 ) *9!*5! Ways of seating
Question 4
IIIIMPPSSSS
n=11
I=4, M =1, P=2, and S =4

MATHEMATICS 4
a) = 11!
4 !∗2 !∗4 ! 1 ! = 34650
b) Since I= 4, then n=8 we get;
= 8 !
1!∗1 !∗2!∗4 ! = 840
c) n=5
= 5!
2! = 60 words
d) = 11 !
4 ! 4 ! 2 ! = 34650 = = 1
34650 = 0.0002886
Question 5
We use two ways to determine the number of times the statement counted and executed i.e.
(n+ 3−1
3 ) = (n+ 2
3 ) times (i+1
2 ) executions of the I, j thus;
( n+2 ) ( n+1 ) n
3 =∑
i=1
n
( i+1
2 )
∑
i=1
n
i = ( n+1 ) n
2
∑
i=1
n
i2= ( n+2 ) ( n+1 ) n
3 - ( n+1 ) n
2
= ( n+1 ) n
6 (2n+1)
( n+ 2
3 )= ∑
i=1
n
( i+1
2 )

MATHEMATICS 5
= ( n+ 2 ) !
3! ( n−1 ) !
( n+2 ) ( n+1 ) (n)
6
= ( 2
2 ) + (3
2 ) +-------+ (n+1
2 )
= ½ (2+6+12+------ (n+1)(n))
= ( n+2 ) ( n+1 ) (n)
6
=1/2 ∑
i=1
n
(i+1 )i
Question 6
X1+ X2+ X3+ X4+ X5+ X6+ X7+ X8+ X9= 27
= X1-1+ X2-1+ X3-1+ X4-1+ X5-1+ X6-1+ X7-1+ X8-1+ X9-1=18
=Consider a= X1-1, b= X2-1, c= X3-1, --------, i= X9-1
=a+b+c+ --------------+i =18
=We therefore know that the number of the solution of the first equation where x1 to x9 cannot
be zero thus equation 2 also cannot be zero. As a result, using the fictious partition method;
(n+ p−1
p−1 ) = (18+ 9−1
9−1 ) where n=18 and p =9
¿ ( 26
8 ) = 26 !
8 !(26−8)!

MATHEMATICS 6
=1562275
Question 7
X1+ X2+ X3+ X4+ X5+ X6+ X7=23
(n+ p−1
p−1 ) = (23+7−1
7−1 ) where n=23 and p =7
¿ ( 26
8 ) = 23 !
6 !(26−4 )!
=10925460
Question 8
∑
j=0
n
(n
j )2j = 3n
By induction, the statement is true when n =0, thus checking on the Left hand side and Right
hand side, we get;
∑
j=0
n
(0
0 )20 =30
This is true since the LHS and the RHS are equal thus holding the claim true to the statement.
∑
j=0
k +1
(k +1
j )20 =3k+1
=3 thus showing that the LHS and the RHS are equal leading to approve of the theorem.
Question 9
A, B, C є U