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MACM 201 Test 1 - Solutions
(1) (15 marks) 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 Solution:
The number of solutions is the same as the number of solutions to
x 1 + x 2 + x 3 + x 4 = 123
0 ≤ x 1 ≤ 65 , 0 ≤ x 2 ≤ 15 , 0 ≤ x 3 ≤ 18 , 0 ≤ x 4 Let c 1 be the condition x 1 ≥ 66, c 2 be x 2 ≥ 16, and c 3 be x 3 ≥ 19.
N (c 1 ) : x 1 + x 2 + x 3 + x 4 = 57, 0 ≤ x 1 , x 2 , x 3 , x 4. So N (c 1 ) =
( 4 + 57 − 1 57
)
N (c 2 ) : x 1 +x 2 +x 3 +x 4 = 107, 0 ≤ x 1 , x 2 , x 3 , x 4. So N (c 2 ) =
( 4 + 107 − 1 107
)
N (c 3 ) : x 1 +x 2 +x 3 +x 4 = 104, 0 ≤ x 1 , x 2 , x 3 , x 4. So N (c 3 ) =
( 4 + 104 − 1 104
)
N( (c 1 c 2 ) : x 1 + x 2 + x 3 + x 4 = 41 , 0 ≤ x 1 , x 2 , x 3 , x 4. So N (c 1 c 2 ) = 4 + 41 − 1 41
)
N( (c 1 c 3 ) : x 1 + x 2 + x 3 + x 4 = 38 , 0 ≤ x 1 , x 2 , x 3 , x 4. So N (c 1 c 3 ) = 4 + 38 − 1 38
)
N( (c 2 c 3 ) : x 1 + x 2 + x 3 + x 4 = 88 , 0 ≤ x 1 , x 2 , x 3 , x 4. So N (c 2 c 3 ) = 4 + 88 − 1 88
)
N( (c 1 , c 2 , c 3 ) : x 1 + x 2 + x 3 + x 4 = 22, 0 ≤ x 1 , x 2 , x 3 , x 4. So N (c 1 , c 2 , c 3 ) = 4 + 22 − 1 22
)
N : x 1 + x 2 + x 3 + x 4 = 123, 0 ≤ x 1 , x 2 , x 3 , x 4. So N =
( 4 + 123 − 1 123
)
Our answer is
N (¯c 1 , ¯c 2 , ¯c 3 ) = N −[N (c 1 )+N (c 2 )+N (c 3 )]+[N (c 1 c 2 )+N (c 1 c 3 )+N (c 2 c 3 )]−N (c 1 c 2 c 3 )
(2) (10 marks) Determine the coefficient of x^13 in the function f (x) where
f (x) = x^2
x
− 4 x^2
)− 5 −
(3 + 5x^8 + x^10 ) 1 + 2x
Solution:
x^2 ( 2 x −^4 x
2
x^2 [ 2 x (1^ −^2 x
] 5
x^7 25 (1 − 2 x^3 )^5
=
x^7 25
∑^ ∞ i=
( 5 + i − 1 i
) 2 ix^3 i
We need the coefficient of x^6 , i.e., i = 2.
(3 + 5x^8 + x^10 ) 1 + 2x
= (3 + 5x^8 + x^10 )(1 − 2 x + 4x^2 − 8 x^3 + · · · + (−1)n 2 nxn^ + · · ·)
The coefficient of x^13 here is 3(−1)2^13 + 5(−1)2^5 − 23
So the answer is; 2 −^3
( 6 2
)
(3) (10 marks) An unlimited supply of red, blue, yellow, and green marbles is available.
(a) (5 marks) 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? What coefficient are we looking for?
Solution:
First of all, if n is odd then the answer is 0 because the number of balls chosen this way has to be even (it’s twice the number of red, blue, and yellow balls chosen). Once the red, blue and yellow balls have been chosen, there is only one way to choose the green balls. So assume n is even; n = 2k. Then the generating function is red (1 + x^2 + x^4 + · · ·)
blue (x^3 + x^5 + · · ·)
yellow (x^7 + x^8 + · · · + x^15 ) = 1 (1 − x^2 )
x^3 (1 − x^2 )
x^7 (1 − x^9 ) (1 − x)
