The Manhattan Cab Problem (Combinations)
In the following activities, students will discover the relationship between Pascal’s
triangle and probability. Students will also see that Pascal’s triangle can be used to
calculate the different combinations and total number of possible outcomes of an
event. These activities are designed to be done by pairs of students.
Grade Level:
5-8
Materials:
Probability and Combinations Activity Sheet
Manhattan Cab Problem Activity Sheet
Pascal’s Triangle (for reference)
Grid paper
Colored pencils
The probability of an event tells how likely it is that the event will occur. In the Coin
Flipping activity of the Patterns in Nature CATE, students will discover that over the
course of many trials, the results of the experimental probability are very close to the
theoretical probability. Pascal’s triangle can be used to calculate the different
combinations which can occur during these experiments, as well as the total number
of possible outcomes.
Have students work in groups of two to four to complete the Probability and
Combinations activity sheet. When they have finished, discuss their findings. Next,
instruct students to complete the Manhattan Cab Problem activity sheet. Discuss the
advantages of using Pascal’s triangle when finding combinations.
SSS
Sunshine State Standards
MA.A.1.2.4
MA.C.3.2.2
MA.C.3.3.2
MA.E.2.2.1
MA.E.2.2.2
MA.E.2.3.1
MA.E.2.3.2
MA.E.3.3.1
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Probability and Combinations
A. Probability
The probability of an event is defined as the ratio of the number of ways the event can happen to
the total possible outcomes. For example, when flipping a coin, the probability of the coin landing
heads up is 1:2, or 50%. Pascal’s triangle can be used to calculate different combinations and the
total number of outcomes of an event. For example, when three coins are tossed in the air, the
possible outcomes of heads and tails are:
3 heads ---> HHH = 1
2 heads and 1 tail ---> HHT, HTH, THH = 3
1 head and 2 tails ---> HTT, THT, TTH = 3
3 tails ---> TTT = 1
1.
In the example, where do the numbers 1, 3, 3, 1 appear in Pascal’s triangle?
___________________
2.
In the example, how many total outcomes are possible? _______
3.
What is the probability of getting 2 heads and 1 tail? _______ 3 heads? ________
4.
List all possible outcomes if four coins were tossed.
5.
How many total outcomes are possible? _________
6.
In how may ways did each of the following outcomes occur?
4 heads = ____;
3 heads, 1 tail = ____;
2 heads, 2 tails = ____;
1 head, 3 tails = ____; 4 tails = ____
7.
Where do these numbers appear in Pascal’s triangle?
8.
What is the pattern for the total number of possible outcomes? ________________________
9.
If ten coins were tossed, in how many ways would each of the following outcomes occur?
10 heads = ________
8 heads, 2 tails = ________
6 heads, 4 tails = _________
4 heads, 6 tails = _________
2 heads, 8 tails = _________
10 tails = _________
9 heads, 1 tail = _________
7 heads, 3 tails = _________
5 heads, 5 tails = _________
3 heads, 7 tails = _________
1 head, 9 tails = __________
10. Why is the number of outcomes for 7 heads, 3 tails the same as the number of outcomes for 3
tails, 7 heads? _______________________________________________________________
_____________________________________________________________________________
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Probability and Combinations
B. The Counting Principle
A school club has five members: three girls, Ann, Beth, and Deb, and two boys, Chris and Eli. If
they want to form a committee of one girl and one boy, how many different committees can be
formed? This problem can be solved by making a list, a tree diagram, or by using the Counting
Principle.
List:
Ann, Chris
Ann, Eli
Beth, Chris
Beth, Eli
Deb, Chris
Deb, Eli
Tree Diagram:
Beth
Ann
Chris
Eli
Chris
Deb
Eli
Chris
Eli
Counting Principle: 3 girls, 2 boys; 3 X 2 = 6 committees
The Counting Principle states, “If one item is to be selected from each of two or more sets, the
total number of possible combinations is the product of the numbers in each set.”
C. Combinations
Someone on the above committee complained that two boys or two girls can do just as good a job
as one girl and one boy. If the committee can have any two members of the club, how many
different committees can now be formed?
List:
Ann, Beth
Ann, Chris
Ann, Deb
Ann, Eli
Beth, Chris
Beth, Deb
Beth, Eli
Chris, Deb
Chris, Eli
Deb, Eli
1.
Why isn’t “Beth, Ann” listed? _________________________________________
2.
How many committees can be formed? _______
3.
Draw a tree diagram to represent each possible committee. (Avoid duplicating committees.)
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Probability and Combinations
A combination is the number of distinct subsets containing r elements that can be selected from a
set S of n elements. In the first committee problem, the club is set S, n is 5, and r is 2. This can
be expressed with the notation
 n
 
r
Since there were 10 combinations of 5 people taken 2 at a time, it can be expressed that
 5
  = 10
 2
For very large sets S, identifying all of the subsets becomes a very tedious process of listing all of
the possibilities. However, the following formula makes it much simpler.
If we define n! (n factorial) to be the product n! = n ⋅ (n − 1) ⋅ (n − 2) ⋅ ... ⋅ 3 ⋅ 2 ⋅ 1, then
n!
 n
 =
 r  r!(n − r )!
5!
5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅1
 5
=
= 10
 =
 2 2!(5 − 2)! 2 ⋅ 1(3 ⋅ 2 ⋅ 1)
4.
Another way to find the number of committees is to use Pascal’s triangle. There are five club
members being taken two at a time. Look at the third term in the fifth row of Pascal’s
triangle. What is it? _____
5.
It appears as if the solutions to these can be found by looking at the (r + 1)th term in the nth
row. Predict how many different committees could be formed if a club of six students wanted
two people on a committee. ______ Verify this with a listing of possible committees. Use
“a, b, c, d, e, and f” as the club members.
6.
Compute each of the following by using the formula and verify the results by using Pascal’s
triangle.
 4
a.  
 2
 6
b.  
 3
Pascal—Sierpinski Connection
 7
c.  
 4
 8
d.  
 6
81
11
e.  
4 
Unit 3
Developed through the National Science Foundation Project Pattern Exploration: Integrating Mathematics and Science for the Middle Grades, 1999
The Manhattan Cab Problem
A taxi driver wants to find the shortest routes between the point of departure to the point of
pickup. There are no one way streets, no diagonal roads, and the driver always travels north or
east, as backtracking wastes time and gas. The map below identifies location B where a customer
is waiting for pickup. The cab driver is located at A. How many different routes (of shortest
mileage) are available for a driver to follow? In this example, the cab driver would travel a
minimum of four blocks, two blocks north and two blocks east, from point A to point B, but he
could arrive at point B in six different ways.
Possible Routes
B
A
B
A
B
A
B
A
B
A
B
A
1.
B
A
Use the grid paper to determine to number of different routes that can be taken from the
point of departure, (A, 0) to each of the following points. Colored pencils may help to show
the different routes.
5
(A,0) = _____ (B,0) = _____ (B,1) = _____ (A,1) = ______
4
(B,2) = _____ (B,3) = _____ (C,1) = _____ (C,2) = ______
3
(A,2) = _____ (D,1) = _____ (C,3) = _____ (D,2) = ______
2
(D,0) = _____ (A,3) = _____ (C,0) = _____ (D,3) = ______
2
1
0
1
A
1
B
C
D
E
F
Write the number of routes at the
intersection, as shown.
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The Manhattan Cab Problem
2.
Examine point (C,3). Which adjacent points must be passed through to arrive at (C,3)?
_______________
3.
What is the relationship between the number of routes to these adjacent points and the
number of routes to (C,3)? ___________________________
4.
Turn the grid so that the point (A,0) is at the top of the paper. Do the numbers look familiar?
_________________________________________________________
5.
If the city were made up of only square sections and the driver wanted to drive from the
extreme southwest corner of the city to the extreme northeast corner of the city, how many
routes could the taxi driver follow in a city where there are exactly:
2 x 2 square blocks? ________
6.
3 x 3 square blocks? ________
Is there a rule that will predict the number of routes in a city of size n x n square blocks?
Hint: Look for these numbers in Pascal’s triangle. ____________________________________
_____________________________________________________________________________
7.
Use this rule or Pascal’s triangle to find the number of routes in a city of 4 x 4 square blocks
___________;
5 x 5 square blocks _________
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Grid Paper
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Other Interesting Features
This activity focuses on two more features of Pascal’s triangle. One feature examines
hidden perfect squares; the other looks at divisibility of numbers on intersecting
diagonals. These activities are designed to be completed in groups of two to four
students.
Grade Level:
5-8
Materials:
Pascal’s Triangle: Diagonals and Rows Reference Sheet
Divisibility Rules Reference Sheet
Hidden Surprises Activity Sheet
Calculators (optional)
Although this CATE includes many features of Pascal’s triangle, other features have
been omitted, mainly because they utilize higher mathematical skills which are
beyond the skills of most of the students in the intended audience. Here are two
more activities that are appropriate for middle level students.
Distribute the Hidden Surprises activity sheet along with copies of Pascal’s triangle
and calculators. Allow students to work through the activities in groups to discover
these additional features. Discuss their findings when the groups are finished.
SSS
Sunshine State Standards
MA.A.1.3.1
MA.A.1.3.2
MA.A.3.2.1
MA.A.3.2.3
MA.A.5.2.1
MA.D.1.2.1
MA.D.1.2.2
MA.D.1.3.1
MA.D.1.3.2
MA.D.2.2.1
MA.D.2.3.1
MA.E.1.2.1
SC.H.2.3.1
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Hidden Surprises
Surprise #1
Work with your group to determine what these surprises are. Be prepared to share your discoveries
with the class.
Examine the intersecting diagonals in the figure below.
Use Pascal’s Triangle: Diagonals and Rows Reference Sheet and Divisibility Rules Reference Sheet to
complete the chart:
3
numbers in the “V” formed by the
intersection of the (n-1)th diagonals
1, 3, 6, 3, 1
n divides all numbers except the outer
“1s” (yes or no)
yes
4
1, 4, 10, 20, 10, 4, 1
no
n
5
6
7
8
9
From the results of the chart, what characteristic must n have in order for it to divide all of the
numbers except the outer 1s in the intersecting diagonals? _____________
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Hidden Surprises
Surprise #2
1.
Look at the second term of row 3. What is it?_____
2.
The numbers 2, 3, 6, 4, 1, and 1 form a hexagon around it. Write the product of these six
numbers in factor form. ________________________________________________________
3.
Write the prime factorization of the product. _______________________________________
4.
What does this prime factorization suggest about the product?_________________________
___________________________________________________________________________
5.
What is the fifth term in row 6? ______
6.
Which six terms surround it? __________________________
7.
Write the product of these six numbers in factor form. _______________________________
8.
Write the prime factorization of the product. _______________________________________
9.
Does this prime factorization have the same characteristic as the one in problem 3?________
10. The numbers 4, 9, 16, 25, 36, 49, and 64 are perfect squares. Find and describe the
characteristics of their prime factorizations. ________________________________________
____________________________________________________________________________
11. Make a conjecture about the product of the six numbers surrounding any term in Pascal’s
triangle. Use a calculator to see if the conjecture is correct. __________________________
____________________________________________________________________________
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Answer Keys for
Miracles of Pascal’s Triangle Unit 3
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Row Properties of Pascal’s Triangle Activity Sheet
Answer Key
1.
2.
3.
4.
5.
6.
two
one
odd
even
3, 7, and 15
2, 4, and 8; yes; 16 is also a power of two; 32 will be the next row, as 32 is the next power of two
after 16
7. The rows which have all odd numbers are always the row before the rows which have all even numbers.
(The row numbers of the rows which have all odd numbers are 2n - 1). Students may find additional
characteristics.
8. All of the numbers in the row except the outer ones are divisible by the prime.
9. The rows are symmetric about their middle number(s).
10. Complete the chart.
Row
Sum of row
Cumulative sum of rows
0
1
1
1
2
1+2=3
2
4
1+2+4=7
3
8
1 + 2 + 4 + 8 = 15
4
16
1 + 2 + 4 + 8 + 16 = 31
5
32
1 + 2 + 4 + 8 + 16 + 32 = 63
6
64
1 + 2 + 4 + 8 + 16 + 32 + 64 = 127
n
2n
2n+1 -1
From the information in the chart, complete the following: The sum of all numbers in any row is 2n,
where n is the row number. The sum of the numbers above row n is 2n+1 - 1.
11.
Row
0
1
2
3
4
5
6
7
8
S1
1
0
0
0
0
0
0
0
0
S2
1
2
2
2
2
2
2
2
2
9
10
n>0
0
0
0
2
2
2
12. If the terms are read as one complete number, 110 is row 0; 111 is row 1; 112 is row 2; 113 is row 3; 114
is row 4.
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Triangular and Tetrahedral Numbers
Answer Key
1.
2.
3.
4.
5.
Row Order
Number of Chips
Sum of Rows
Triangular Number
1
2
3
4
5
1
2
3
4
5
1
1+2=3
1+2+3=6
1 + 2 + 3 + 4 = 10
1 + 2 + 3 + 4 + 5 = 15
1
3
6
10
15
yes; in the second diagonal
3× 4
=6
yes;
2
4×5
= 10
yes;
2
n(n + 1)
yes;
2
Layers Needed
6.
1
2
3
4
Spheres in
Bottom Layer
1
3
6
10
Total Number
Of Spheres
1
1+3=4
10
20
7.
yes; they make up the third diagonal
8.
yes; each layer was comprised of the next triangular number
9.
Each tetrahedral number is the sum of the triangular numbers in the diagonal above it.
10. yes
11. 5005 - ( 1 + 9 + 45 ) = 4950 Subtract the sum of the numbers above the number being
summed from the number below and in the opposite direction of the number being summed.
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Fibonacci Numbers Activity Sheet
Answer Key
1.
First Day of
Jan
Feb
Mar
Apr
May Jun
Jul
Aug Sep
Oct
Nov Dec
Adult Pairs
1
1
2
3
5
8
13
21
34
55
89
144
Newborn
Pairs
0
1
1
2
3
5
8
13
21
34
55
89
1
2
3
5
8
13
21
34
55
89
144
233
Total Pairs
F7 = 13
F10 = 55
F15 = 610
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Probability and Combinations
Answer Key
A. Probability
1.
2.
3.
4.
The numbers are row three of Pascal’s triangle.
8
3/8; 1/8
4 heads ---> HHHH
3 heads, 1 tail ---> HHHT, HTHH, HHTH, THHH
2 heads, 2 tails ---> HHTT, HTHT, HTTH, THHT, THTH, TTHH
1 head, 3 tails ---> HTTT, THTT, TTHT, TTTH
4 tails ---> TTTT
5. 1; 4; 6; 4; 1
6. They appear in row 4.
7. 16
8. 2n
9. 10 heads = 1
9 heads, 1 tail = 10
8 heads, 2 tails = 45
7 heads, 3 tails = 120
6 heads, 4 tails = 210 5 heads, 5 tails = 252
4 heads, 6 tails = 210 3 heads, 7 tails = 120
2 heads, 8 tails = 45
1 head, 9 tails = 10
10 tails = 1
10. Pascal’s triangle is symmetric. Seven of one outcome and three of another are the same.
B. The Counting Principle
no answers needed
C. Combinations
1.
2.
3.
4.
5.
Beth, Ann is the same committee as Ann, Beth
10
10
15
ab
ac
ad
ae
af
bc
bd
be
bf
cd
ce
cf
de
df
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Probability and Combinations
Answer Key
6.
a.
6
4!
4 ⋅ 3 ⋅2 ⋅1
=
=6
2! ( 4 − 2)! 2 ⋅ 1(2 ⋅ 1)
b.
20
6!
6 ⋅5 ⋅ 4 ⋅ 3 ⋅2 ⋅1
=
= 20
3! (6 − 3)! 3 ⋅ 2 ⋅ 1(3 ⋅ 2 ⋅ 1)
c.
35
7!
7 ⋅6 ⋅5 ⋅ 4 ⋅ 3 ⋅2 ⋅1
=
= 35
4! (7 − 4)! 4 ⋅ 3 ⋅ 2 ⋅ 1(3 ⋅ 2 ⋅ 1)
d.
28
8!
8 ⋅7 ⋅6 ⋅5 ⋅ 4 ⋅ 3 ⋅2 ⋅1
=
= 28
6! (8 − 6)! 6 ⋅ 5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅ 1(2 ⋅ 1)
e.
330
11!
11 ⋅ 10 ⋅ 9 ⋅ 8 ⋅ 7 ⋅ 6 ⋅ 5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅ 1
=
= 330
4! (11 − 4)!
4 ⋅ 3 ⋅ 2 ⋅ 1(7 ⋅ 6 ⋅ 5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅ 1)
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The Manhattan Cab Problem
Answer Key
1.
(A,0) = 1
(B,2) = 3
(A,2) = 1
(D,0) = 1
(B,0) = 1
(B,3) = 4
(D,1) = 4
(A,3) = 1
2.
(B,3) (C,2)
3.
4 + 6 = 10
4.
They form Pascal’s triangle.
5.
6; 20
6.
The number will be the (n + 1)th term of the (2n)th row of Pascal’s triangle.
OR
The number will be the middle term in an even row—find the row by doubling the side of the
grid.
7.
70; 252
Pascal—Sierpinski Connection
(B,1) = 2
(C,1) = 3
(C,3) = 10
(C,0) = 1
(A,1) = 1
(C,2) = 6
(D,2) = 10
(D,3) = 20
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Hidden Surprises
Answer Key
Surprise #1
3
numbers in the “V” formed by the
intersection of the (n-1)th diagonals
1, 3, 6, 3, 1
n divides all numbers except the outer
“1s” (yes or no)
yes
4
1, 4, 10, 20, 10, 4, 1
no
5
1, 5, 15, 35, 70, 35, 15, 1
yes
6
1, 6, 21, 56, 126, 252, 126, 56, 21, 6, 1
no
7
1, 7, 28, 84, 210, 462, 924, 462, 210,
84, 28, 7, 1
1, 8, 36, 330, 792, 1716, 3432, 1716,
792, 330, 120, 36, 8, 1
1, 9, 45, 165, 495, 1287, 3003, 6435,
12870, 6435, 3003, 1287, 495, 165, 45,
9, 1
yes
n
8
9
no
no
Surprise #1: It appears as if n must be prime in order for it to divide all of the numbers in the
(n - 1)th diagonals except the outer 1s.
Surprise #2:
1. 3
2. 2 ⋅ 3 ⋅ 2 ⋅ 3 ⋅ 2 ⋅ 2 ⋅ 1
3. 2 4 ⋅ 32
4. Answers will vary. The exponents are all 2s and 4s. There is an even number of factors. The
product is a perfect square.
5. 15
6. 5, 6, 21, 35, 20, and 10
7. 5 ⋅ 2 ⋅ 3 ⋅ 3 ⋅ 7 ⋅ 5 ⋅ 7 ⋅ 2 ⋅ 2 ⋅ 5 ⋅ 2 ⋅ 5
8. 2 4 ⋅ 32 ⋅ 54 ⋅ 72
9. Yes, all of the exponents are even (2s and 4s).
10. 4 = 2 2
9 = 32
16 = 2 4
25 = 52
36 = 2 2 ⋅ 32
49 = 72
64 = 2 6
All of their exponents are even.
11. The product of the six numbers surrounding any term in Pascal’s triangle will always be a perfect
square.
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