Integer Solutions - Discrete Mathematics - Exam, Exams of Discrete Mathematics

This is the Exam of Discrete Mathematics which includes Recurrence Relation, Space Is Available, Answer, Number of Ways, Sum of Odd Integers, By Hand, Solution, Various Walks, Provided etc. Key important points are: Integer Solutions, Inclusion and Exclusion, Coefficient, Function, Unlimited Supply, Green Marbles, Generating Function, Even Nu, Odd Number, Same Number

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2012/2013

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MACM 201 Test 1
February 6, 2006. 50 minutes
Total marks: 45. Marks are indicated by ( ).
Nc1,¯c2¯c3· · · ¯ct) = N[N(c1) + N(c2) + · · · +N(ct)]
+[N(c1c2) + N(c1c3) + · · · +N(c1ct) + · · · +N(c2c3) + · · · +N(ct1ct)]
[N(c1c2c3) + N(c1c2c4) + · · · +N(c1c2ct) + N(c1c3c4) + · · ·
+N(c1c3ct) + · · · +N(ct2ct1ct)] + · · · + (1)tN(c1c2c3· · · ct)
=S0S1+S2S3+· · · + (1)tSt
Em=Sm m+ 1
1!Sm+1 + m+ 2
2!Sm+2 · · · + (1)tm t
tm!St
Lm=Sm m
m1!Sm+1 + m+ 1
m1!Sm+2 · · · + (1)tm t1
m1!St
If nZ+,
n
r!= n+r1
r!
For all m,n Z+, a R,
(1 + x)n= n
0!+ n
1!x+ n
2!x2+· · · + n
n!xn
(1 xn+1)
(1 x)= 1 + x+x2+x3+·· · +xn
1
(1 x)= 1 + x+x2+x3·· · =
X
i=0
xi
1
(1 + x)n= n
0!+ n
1!x+ n
2!x2+· · ·
=
X
i=0 n
i!xi
= 1 + (1) n+ 1 1
i!x+ (1)2 n+ 2 1
2!x2+· · ·
=
X
i=0
(1)i n+i1
i!xi
1
(1 x)n= n
0!+ n
1!(x) + n
2!(x)2+· · ·
=
X
i=0 n
i!(x)i
= 1 + (1) n+ 1 1
i!(x)+(1)2 n+ 2 1
2!(x)2+· · ·
=
X
i=0 n+i1
i!xi
1
pf3
pf4
pf5

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MACM 201 Test 1

February 6, 2006. 50 minutes

Total marks: 45. Marks are indicated by ( ).

N (¯c 1 , ¯c 2 c¯ 3 · · · ¯ct) = N − [N (c 1 ) + N (c 2 ) + · · · + N (ct)]

+[N (c 1 c 2 ) + N (c 1 c 3 ) + · · · + N (c 1 ct) + · · · + N (c 2 c 3 ) + · · · + N (ct− 1 ct)] −[N (c 1 c 2 c 3 ) + N (c 1 c 2 c 4 ) + · · · + N (c 1 c 2 ct) + N (c 1 c 3 c 4 ) + · · · +N (c 1 c 3 ct) + · · · + N (ct− 2 ct− 1 ct)] + · · · + (−1)tN (c 1 c 2 c 3 · · · ct) = S 0 − S 1 + S 2 − S 3 + · · · + (−1)tSt

Em = Sm −

( m + 1 1

) Sm+1 +

( m + 2 2

) Sm+2 − · · · + (−1)t−m

( t t − m

) St

Lm = Sm −

( m m − 1

) Sm+1 +

( m + 1 m − 1

) Sm+2 − · · · + (−1)t−m

( t − 1 m − 1

) St

If n ∈ Z+, (^) ( −n r

)

( n + r − 1 r

)

For all m, n ∈ Z+, a ∈ R,

(1 + x)n^ =

( n 0

)

( n 1

) x +

( n 2

) x^2 + · · · +

( n n

) xn

(1 − xn+1) (1 − x)

= 1 + x + x^2 + x^3 + · · · + xn

1 (1 − x)

= 1 + x + x^2 + x^3 · · · =

∑^ ∞

i=

xi

(1 + x)n^

( −n 0

)

( −n 1

) x +

( −n 2

) x^2 + · · ·

∑^ ∞

i=

( −n i

) xi

( n + 1 − 1 i

) x + (−1)^2

( n + 2 − 1 2

) x^2 + · · ·

∑^ ∞

i=

(−1)i

( n + i − 1 i

) xi

(1 − x)n^

( −n 0

)

( −n 1

) (−x) +

( −n 2

) (−x)^2 + · · ·

∑^ ∞

i=

( −n i

) (−x)i

( n + 1 − 1 i

) (−x) + (−1)^2

( n + 2 − 1 2

) (−x)^2 + · · ·

∑^ ∞

i=

( n + i − 1 i

) xi

(1) (15) Do not use generating functions to solve this problem. Use the

Principle of Inclusion and Exclusion (Chapter 8).

How many integer solutions are there to

x 1 + x 2 + x 3 + x 4 = 120

0 ≤ x 1 ≤ 65 , − 5 ≤ x 2 ≤ 10 , 2 ≤ x 3 ≤ 20 , 0 ≤ x 4

(3) (10) An unlimited supply of red, blue, yellow, and green marbles is available.

(a) Write down the generating function that will answer this problem: How

many ways can n marbles be selected so that there are an even number of red

marbles, an odd number of blue marbles, at least 7 but no more than 15 yellow

marbles, and the same number of green marbles as the sum of the red, blue,

and yellow ones? (Remember, 0 is an even number.) What coefficient are we

looking for?

Simplify your answer using the formula on the front page.

(b) Determine the number of ways to select the marbles if n = 22. Describe

how many of each color are selected for these choices.

(c) How many ways are there to select the marbles if n = 21?

(4) (10) Consider the following recurrence relation;

6 an+2 + an+1 − an = 0, n ≥ 0 , a 0 = 2, a 1 = − 3

(a) Write out the terms a 1 , a 1 , a 2 , a 3.

(b) Solve the recurrence recurrence relation and check your answer with the

value of a 3 you found in part (a). What is a 100?