Download Solutions - School of Mathematics and Statistics, University of Sydney

MATH1004
Summer School, 2017
Discrete Mathematics
Solutions to Tutorial 1
1. The first few rows of Pascal’s triangle are written down below. Recall that each entry is
equal to the sum of the two numbers immediately above it.
n = 0:
1
n = 1:
1
n = 2:
1
n = 3:
2
1
n = 4:
1
3
1
4
1
3
6
1
4
1
(a) Write down the next two rows.
Solution: The next two rows are:
n = 5:
n = 6:
1
1
5
6
10
15
10
20
5
15
1
10
1
(b) Add up the numbers in each row. Do you notice a pattern? Guess what the sum
of the entries in the nth row should be.
Solution: The sums of the first 6 rows are:
1
1+1
1+2+1
1+3+3+1
1+4+6+4+1
1 + 5 + 10 + 10 + 5 + 1
1 + 6 + 15 + 20 + 15 + 6 + 1
=
=
=
=
=
=
=
1
2
4
8
16
32
64.
These numbers are increasing powers of 2. A reasonable guess is that the entries of the
nth row of Pascal’s triangle add up to 2n .
(c) Now add up the numbers in each row, but this time after inserting alternating +
and − signs in front of each entry. For example, you should calculate 1 − 3 + 3 − 1
for the third row. What do you notice?
Solution: Apart from the 0th row, the sum of the entries of each row with alternating +
and − signs is equal to 0. One may plausibly conjecture that this is true for all subsequent
rows.
2. A snack shop sells ice cream in the following three flavours: avocado, banana and chocolate. You may order an ice cream cone with up to three scoops, each scoop having a
different flavour.
c 2014 The University of Sydney
Copyright 1 http://www.maths.usyd.edu.au/u/UG/SS/SS1004/
(a) List all possible orders you make with exactly two scoops. How many are there?
Solution: There are 3 possible orders: avocado + banana; avocado + chocolate; banana
+ chocholate.
(b) What about with exactly one scoop? Three scoops? Zero scoops?
Solution: For one scoop there are 3 possible orders, corresponding to the 3 individual
flavours. There is only one order containing 3 scoops: avocado + banana + chocolate.
For zero scoops, there is one possible order: an empty cone!
Suppose the shop introduces a fourth flavour: dragonfruit. You may now order a cone
with up to four different flavours.
(c) How many possible orders with exactly two scoops are there now?
Solution: There are 6 possible orders: avocado + banana; avocado + chocolate; avocado
+ dragonfruit; banana + chocholate; banana + dragonfruit; chocolate + dragonfruit.
(d) What about with exactly 3 scoops? Can you figure this out without listing all the
possibilities? Do the same for exactly 0, 1 or 4 scoops.
Solution: To order a cone with 3 scoops, you must choose exactly one flavour to leave
out and hence there are 4 possible orders. There are 4 possible orders with one scoop out
of 4 flavours. Finally, there is exactly one way to order an empty cone; and likewise for a
cone with all 4 flavours.
(e) Now compare your answers with rows 3 and 4 of Pascal’s triangle. Do you see a
connection? Guess the number of different orders you can make with exactly 2
scoops from 5 possible flavours.
Solution: When there are 3 possible flavours to choose from, the number of orders with
exactly 0, 1, 2, and 3 scoops respectively are 1, 3, 3, and 1. With 4 possible flavours,
the number of orders with exactly 0, 1, 2, 3, and 4 scoops respectively are 1, 4, 6, 4, and
1. These numbers coincide exactly with the entries of the 3rd and 4th rows of Pascal’s
triangle. A plausible guess is that the number of orders consisting of r scoops from n
possible flavours is equal to the (r + 1)th entry of the nth row of Pascal’s triangle. Using
this proposed rule, we may guess that there are 10 ways to choose exactly 2 out of 5
flavours. [You should verify this!]
3. We now investigate expansions of (x + 1)n for various values of n ≥ 0. Recall that
(x + 1)2 = (x + 1)(x + 1) = x2 + 2x + 1.
(a) Expand (x + 1)3 .
Solution: (x + 1)3 = x3 + 3x2 + 3x + 1.
(b) Expand (x + 1)4 . [Hint: multiply your previous answer by (x + 1).]
Solution: (x + 1)4 = x4 + 4x3 + 6x2 + 4x + 1.
(c) Guess a connection between the coefficients in the expansions of (x+1)n and Pascal’s
triangle. [It may be helpful to observe that (x + 1)0 = 1 and (x + 1)1 = x + 1.]
Solution: The coefficients in the expansion of (x + 1)n (written in order of decreasing
powers of x) coincide with the entries of the nth row of Pascal’s triangle.
(d) Using your guess, write down the expansions of (x + 1)5 and (x + 1)6 .
2
Solution: Applying the guess to rows 5 and 6 from Pascal’s triangle obtained from
Question 1(a), we have:
(x + 1)5 = x5 + 5x4 + 10x3 + 10x2 + 5x + 1 and
(x + 1)6 = x6 + 6x5 + 15x4 + 20x3 + 15x2 + 6x + 1.
(e) Now substitute the values x = 1 or x = −1 into your expansions. Do these results
agree with your answers for parts (b) and (c) of Question 1?
Solution:
Substituting x = 1 into the expansion of (x + 1)n gives the sum of its
coefficients. The above guess suggests that this is equal to the sum of the entries of row
n of Pascal’s triangle. Indeed, since (1 + 1)n = 2n , this is consistent with our observed
pattern in Question 1 (b).
Similarly, substituting x = −1 into the expansion of (x + 1)n gives the alternating sum of
its coefficients. We observed that for n ≥ 1, the alternating sum of the nth row of Pascal’s
triangle is 0. Since (−1 + 1)n = 0n = 0, this is consistent with our observed pattern in
Question 1.
3