Additional Homework Problems Math 3000 Fall 2006
2. Suppose |A| = 24, |B| = 21, |A ∪ B| = 37, |A  C| = 11, |B – C| = 10 and |C – B| = 12. Find
a) |A  B|
b) |A - B|
c) |B – A|
d) |B ∪ C|
e) |C|
f) |A ∪ C|
Solutions:
a) Let |A  B| = x. Then 24 – x + 21 – x + x = 37. So, x = 8.
b) 24 – x = 16.
c) 21 – x = 13.
d) Let |B  C| = y. Then 10 + y = 21, so y = 11. |B ∪ C| = 10 + 11 + 12 = 33.
e) 12 + y = 23.
f) 13 + 11 + 12 = 36.
4. Of the four teams in a softball league, one team has four pitchers and the other teams have
three each. Give the counting rules that apply to determine each of the following.
(a) The number of possible selections of pitchers for an all-star team, if exactly four pitchers are to
be chosen.
(b) The number of possible selections if one pitcher is to be chosen from each team.
(c) The number of possible selections of four pitchers, if exactly two of the five left-handed pitchers
in the league must be selected.
(d)The number of possible orders in which the four pitchers, once they are selected, can appear
(one at a time) in the all-star game.
Solutions:
a) Combinations. (
)
b) Product rule: 4 x 3 x 3 x 3
c) Combinations and Product Rule: ( ) ( )
d) Product rule: 4 x 3 x 2 x 1
6. (a) If you have 10 left shoes and 9 right shoes and do not care whether they match, how many
“pairs” of shoes can you select?
(b) A cafeteria has 3 meat selections, 2 vegetable selections, and 4 dessert selections for a given
meal. If a meal consists of one meat, one vegetable and one dessert, how many different meals
could be constructed?
(c)There are 3 roads from Abbottville to Bakerstown, 4 roads from Bakerstown to Cadez, and 5
roads from Cadez to Detour Village. How many different routes are there from Abbottville through
Bakerstown and then Cadez to Detour Village?
Solutions:
a) 90 if a “pair” consists of a left and a right shoe. (
)
if any pair will do.
b) 3 x 2 x 4 = 24.
c) 3 x 4 x 5 = 60.
10. Find the number of ways seven school children can line up to board a school bus.
Solution: 7! = 7 x 6 x 5 x 4 x 3 x 2 x 1.
11.Suppose the seven children of exercise 10 (above) are three girls and four boys. Find the
number of ways they could line up subject to these conditions.
(a) The three girls are first in line.
(b) The three girls are together in line.
(c) The four boys are together in line.
(d) No two boys are together.
Solutions:
a) 3 x 2 x 1 x 4 x 3 x 2 x 1 = 144.
b) 5(144). Counts 3GBBBB, B3GBBB, BB3GBB, BBB3GB, BBBB3G.
c) 4(144). Counts 4BGGG, G4BGG, GG4BG, GGG4B.
d) 144. Must be BGBGBGB.
14.From a second-grade class of 11 boys and 8 girls, 3 are selected for flag duty.
(a) How many selections are possible?
(b) How many of these selections have exactly two boys?
(c) Exactly one boy?
Solutions:
a) (
)
b) (
)( )
c) (
)( )
17.Find
(a) (a + b)6
(b) (a + 2b)4
(c) the coefficient of a3b10 in the expansion of (a + b)13
(d) the coefficient of a2b10 in the expansion of (a + 2b)12
Solutions:
a) a6 + 6 a5b + 15a4b2 + 20a3b3 + 15 a2b4 + 6 ab5 + b6.
b) a4 + 8a3b + 24a2b2 + 32ab3 + 16b4.
c) (
)
d)
(
)
18.(a) Prove combinatorially that if n is odd, then the number of ways to select an even number of
objects from n is equal to the number of ways to select an odd number of objects.
(b) Give a combinatorial proof of Vandermonde's identity: for positive integers m and n, and r an
integer such that 0 ≤ r ≤ n + m,
(
(c) Prove that (
Solutions:
)
)
(
( )( )
)
(
( )(
)
)
( )(
)
( )( )
a) Since n is odd, let n = 2k + 1 for some integer k. The number of ways of selecting an even
number of objects from n is given by ( ) for some even number t. But, choosing t objects is
the same as not choosing n – t objects, and n – t is an odd number. As t varies through all
the even numbers less than n, n – t will run through all the odd numbers.
b) Let there be n red balls and m blue balls. We want to pick r of these balls (ignoring colors).
The number of ways to do this is (
) Now each selection consists of say x red balls
and r-x blue balls. The number of ways to select these is ( ) (
) Letting x vary from 0
to r will give all possible selections which we add by the addition principle.
c) Using the identity, ( )
also have (
(
)
)
(
(
(
)
(
) we get (
)
(
)
) Now using the first relation again, (
) This gives the required equation.
(
) Now we
)
(
)