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SIMON FRASER UNIVERSITY
DEPARTMENT OF MATHEMATICS
Midterm 1 – Solutions
MACM 201 Summer 2007 Instructor: Robert ˇS´amal June 6, 2007, 12:30 – 13:
1. Let S be a (finite) set, c 1 ,... , ct conditions on the elements of S.
[1] (a) Suppose t = 3. State the Principle of Inclusion and Exclusion, and explain the used notation. Don’t use ‘... ’, nor ‘
N( c 1 c 2 c 3 ) = |S| −
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 ) Here (as usually), N(ci) is the number of elements of S satisfying ci, N(cicj ) the number of elements satisfying ci and cj , and N(c 1 c 2 c 3 ) the number of elements satisfying all three conditions. Finally, N( c 1 c 2 c 3 ) is the number of elements satisfying neither of them.
[3] (b) Let t = 3 again, and let S be as on the figure (both full and empty points count). Fill in the following table:
c 1 c 2
c 3
N(c 1 ) N(c 2 ) N(c 3 ) N(c 1 c 2 ) N( c 2 c 3 ) S 1
9 9 5 6 2 23
[1] (c) With the same set S as in the previous part, how many times is the element of S denoted by empty circle counted in the sum S 2 =
1 ≤i<j≤t N(cicj^ )?
3 times
Decide if the following formulas are true or false, and if false, correct them!
[1] (d) N( c 1 ) = N − N( c 1 ) True.
[1] (e) N( c 1 c 2 ) = N( c 2 ) − N( c 1 c 2 ) True.
[1] (f) N( c 1 c 2 ) = N( c 2 ) − N( c 1 c 2 ) False. Possible corrections: N( c 1 c 2 ) = N( c 1 ) − N( c 1 c 2 ) N( c 1 c 2 ) = N( c 2 ) − N( c 1 c 2 )
[2] (g) N( c 1 c 2 ) = N − N( c 1 c 2 c 3 ) − N( c 1 c 2 c 3 ) False. Possible corrections include: N( c 1 or c 2 ) = N − N( c 1 c 2 c 3 ) − N( c 1 c 2 c 3 ) N( c 1 c 2 ) = N − (N( c 1 ) + N( c 2 )) + (N( c 1 c 2 c 3 ) + N( c 1 c 2 c 3 )) N( c 1 c 2 ) = N − N( c 1 c 2 c 3 ) − N( c 1 c 2 c 3 ) − N( c 1 c 2 ) − N( c 1 c 2 )
[10] 3. Two giants, three witches, and seven dwarfs are going to take a group photograph, that
is they need to arrange in a single row. To make the photograph look interesting, they want that
- the two giants don’t stand next to each other,
- the three witches don’t stand all together, and
- the seven dwarfs don’t stand all together. For example, the following is a proper arrangement: G 2 D 7 D 2 D 3 W 3 D 5 D 6 D 4 D 1 W 2 W 1 G 1. How many different photographs can they make in this way?
The set S will consist of all possible arrangements of the creatures, that is of all permutations of 2 + 3 + 7 = 12 objects. (We consider the creatures of the same ‘type’ to be distinguishable. If we suppose they are not, we get another problem, but the solution is similar.) The conditions will be
- c 1 : the two giants do stand next to each other,
- c 2 : the three witches do stand all together,
- c 3 : the seven dwarfs do stand all together.
Let’s find the numbers now. As a typical case, consider N(c 2 c 3 ) first. We know that all witches and all dwarfs stand together: that is we are looking at permutations of 4 ‘objects’ (two dwarfs, a group of witches and a group of dwarfs), and we also have to consider the permutations of the dwarfs and of the witches. So, we get N(c 2 c 3 ) = 4! · 3! · 7!. Let’s go systematic now.
- S 0 = 12!
- S 1 = N(c 1 ) + N(c 2 ) + N(c 3 ), where
- N(c 1 ) = 11! · 2! (we make the two giants into one group, which makes for 11 ‘objects’ to permute).
- N(c 2 ) = 10! · 3! (we make the three witches into one group)
- N(c 3 ) = 6! · 7! (we make the dwarfs into one group)
- S 2 = N(c 1 c 2 ) + N(c 1 c 3 ) + N(c 2 c 3 ), where
- N(c 1 c 2 ) = 9! · 2! · 3!
- N(c 1 c 3 ) = 5! · 2! · 7!
- N(c 2 c 3 ) = 4! · 3! · 7!
- S 3 = N(c 1 c 2 c 3 ) = 3! · 2! · 3! · 7!
We are ready for the final answer. The number of proper arrangements is
N( c 1 c 2 c 3 ) = S 0 −S 1 +S 2 −S 3 = 12!−(11!·2!+10!·3!+6!·7!)+(9!·2!·3!+5!·2!·7!+4!·3!·7!)−3!·2!·3!·7!
To get the idea, this equals 379 693 440 (while 12! = 479 001 600), but you were not supposed to enumerate it.
5. Find
∑n k=1 2 kk^ be means of recurrence relations. (Use the back of the previous page if more space is needed.)
[2] (a) Put an =
∑n k=1 2 kk^ and write a recurrence relation for^ an. an = an− 1 + 2 nn for n ≥ 1.
[2] (b) Find a 1 , a 2 , and a 3 ‘by hand’. a 1 = 1/ 21 = 1/ 2 a 2 = 1/ 21 + 2/ 22 = 1 a 2 = 1/ 21 + 2/ 22 + 3/ 23 = 11/ 8
[5] (c) Solve the recurrence relation found in (a).
- The homogeneous solution is a( nh )= A (A is any constant).
- We will look for a particular solution in form a( np )= (Bn + C)/ 2 n^ (B, C suitable constants). Plugging in the recurrence relation we obtain Bn + C 2 n^ −^
B(n − 1) + C 2 n−^1 =^
n 2 n^. and after some algebra we get −B · n + (2B − C) = n A simple (in fact, the only) way how to satify this equation for every n is solving a system of two equations
−B · n = n 2 B − C = 0
This gives us B = − 1 , C = − 2. We’ve found a( np )= −(n + 2)/ 2 n.
- The general solution is an = a( nh )+ a( np )= A − (n + 2)/ 2 n. To determine A we let n = 1: 1 /2 = a 1 = A − 3 / 2 , hence A = 2. Altogether we have an = 2 − n^2 + 2n (n ≥ 0).
[1] (d) Verify the solution for n = 1, 2 , 3. a 1 = 2 − 3 / 21 = 1/ 2 a 2 = 2 − 4 / 22 = 1 a 3 = 2 − 5 / 23 = 11/ 8 , so it works.
