Let’s Start With A Product Rule Example
Example 1:
Lisa has 3 skirts (Brown, Tan, and Green), 5 shirts (White, Pink, Yellow, Red, and
Striped), and 2 pairs of shoes (Heals and Flats). Assume that an outfit consists of a skirt,
a shirt, and a pair of shoes. How many outfits can Lisa make?
• Ask yourself what it is you’re counting. List some examples of the things you’re trying
to count. If you can’t do this step STOP! Reread the question, pause for a minute to
think about it, or ask Jackie a clarification question.
We’re counting the outfits that Lisa can make. Examples are Brown skirt, Red shit,
and Heals (abbreviated BRH) or Tan skirt, Red shirt, and Flats (abbreviated TRF)
• Ask yourself if you can break each of the things your counting down into smaller pieces.
For this example we can. Every outfit is made up of a skirt, a shirt and a pair of shoes.
Skirt Shirt Pair of Shoes
• If possible, list the choices or possibilities for every piece
For this example we can:
choices
Skirt
Brown, Tan, Green
Shirt
White, Pink, Yellow, Red, Striped
Pair of Shoes
Heals, Flats
• List the number of choices for every piece. Make sure that when you write down a
number of choices, there are always exactly that many choices for that piece.
For this example, since we were able to write down exactly what the choices were for
each piece, we just need to count them.
choices
# choices
Skirt
Brown, Tan, Green
3 choices
Shirt
White, Pink, Yellow, Red, Striped
5 choices
Pair of Shoes
Heals, Flats
2 choices
• If you’ve set everything up correctly, then then multiply then number of choices for
each step to get the answer.
So there are 30 outfits Lisa can make.
3 × 5 × 2 = 30
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What Does it Mean?
With the product rule, we counted all 30 entries listed in the tree diagram below.
*
B
W
P
Y
T
R
S
HF HF HF HF HF
W
P
G
Y
R
S
W
HF HF HF HF HF
P
Y
R
HF HF HF HF HF
Recall that to interpretative a tree diagram we look at all the ‘leaves’ (which are along the
bottom here), we need to look at the ‘Path’ from the ∗ to that leaf. So the first leaf represents
BWH, the second leaf represents BWF, and so on. So here’s here’s the list of the 30 outfits
Lisa can make.
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
Brown
Brown
Brown
Brown
Brown
Brown
Brown
Brown
Brown
Brown
Tan
Tan
Tan
Tan
Tan
skirt White Shirt Heals
skirt White Shirt Flats
skirt
Pink Shirt Heals
skirt
Pink Shirt Flats
skirt Yellow Shirt Heals
skirt Yellow Shirt Flats
skirt
Red Shirt Heals
skirt
Red Shirt Flats
skirt Striped Shirt Heals
skirt Striped Shirt Flats
skirt White Shirt Heals
skirt White Shirt Flats
skirt
Pink Shirt Heals
skirt
Pink Shirt Flats
skirt Yellow Shirt Heals
16.
17.
18.
19.
20.
21.
22.
23.
24.
25.
26.
27.
28.
29.
30.
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Tan
Tan
Tan
Tan
Tan
Green
Green
Green
Green
Green
Green
Green
Green
Green
Green
skirt
skirt
skirt
skirt
skirt
skirt
skirt
skirt
skirt
skirt
skirt
skirt
skirt
skirt
skirt
S
Yellow
Red
Red
Striped
Striped
White
White
Pink
Pink
Yellow
Yellow
Red
Red
Striped
Striped
Shirt
Shirt
Shirt
Shirt
Shirt
Shirt
Shirt
Shirt
Shirt
Shirt
Shirt
Shirt
Shirt
Shirt
Shirt
Flats
Heals
Flats
Heals
Flats
Heals
Flats
Heals
Flats
Heals
Flats
Heals
Flats
Heals
Flats
Example 2:
How many odd 6 digit numbers are there? Recall that numbers that end in 0,2,4,6, or 8 are
even, and numbers than end in 1,3,5,7, or 9 are odd. Remember also that a ‘6 digit number ’
can’t start with a zero: 019,474 is a 5 digit number, but 319,474 is a 6 digit number.
• Ask yourself what it is you’re counting. List some examples of the things you’re trying
to count. If you can’t do this step STOP! Reread the question, pause for a minute to
think about it, or ask Jackie a clarification question.
In this example we’re counting all of the odd 6 digit numbers. Some examples of odd
6 digit numbers are 234567, 777777, 868685, and 906503.
• Ask yourself if you can break each of the things your counting down into smaller pieces.
For this example we can. Every 6 digit number is made up of:
First Digit Second Digit Third Digit Forth Digit Fifth Digit Sixth Digit
• If possible, list the choices or possibilities for every piece
For this example we can:
First Digit Second Digit Third Digit Forth Digit Fifth Digit Sixth Digit
choices
1,2,3,4,
5,6,7,8,9
0,1,2,3,4,
5,6,7,8,9
0,1,2,3,4,
5,6,7,8,9
0,1,2,3,4,
5,6,7,8,9
0,1,2,3,4,
5,6,7,8,9
1,3,5,7,9
• List the number of choices for every piece. Make sure that when you write down a
number of choices, there are always exactly that many choices for that piece.
For this example, since we were able to write down exactly what the choices were for
each piece, we just need to count them.
First Digit Second Digit Third Digit Forth Digit Fifth Digit Sixth Digit
choices
1,2,3,4,
5,6,7,8,9
0,1,2,3,4,
5,6,7,8,9
0,1,2,3,4,
5,6,7,8,9
0,1,2,3,4,
5,6,7,8,9
0,1,2,3,4,
5,6,7,8,9
1,3,5,7,9
# choices
9
10
10
10
10
5
• If you’ve set everything up correctly, then then multiply then number of choices for
each step to get the answer.
So there are 450, 000 odd 6 digit numbers.
9 × 10 × 10 × 10 × 10 × 5 = 450, 000
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An Example Where We Can’t ONLY Use The Product Rule
Example 3:
At a certain middle school the 7th graders have ‘electives’ during third and forth period.
During third period Band, Drawing, Gym, and Choir are offered, and during forth period
Sculpture, Photography, Yearbook, Wood shop, and Marching band. In the beginning of
the school year the 7th graders are asked to choose an ‘Elective schedule’ where they pick
one elective for each period. The Marching band director requires that any student who
takes Marching Band takes either Band or Choir as their other elective. How many
different ‘Elective schedules’ are possible.
What happens if we try to only use the product rule?
We notice that we can break an ‘Elective schedule’ into a third period class and a forth
period class:
choices
Third Period Class
B, D, G, C
Forth Period Class
S, P, Y W, M???
Should we include Marching Band?
Including M, would give us the following 4 × 5 = 20 choices
*
B
D
G
C
SPYWM SPYWM SPYWM SPYWM
But DM and GM are NOT allowed, so 4 × 5 = 20 was NOT the correct way to count.
Not Including M, would give us the following 4 × 4 = 16 choices
*
B
D
G
C
SPYW SPYW SPYW SPYW
But this list misses BM and CM are possible choices, so 4 × 4 = 16 was NOT the correct
way to count.
5
We Better Try Something Else...
Let’s divide the thing we’re trying to count (‘Elective schedules’) into disjoint cases. Our
Disjoin cases will be:
• ‘Elective schedule’ with Marching band 4th period
• ‘Elective schedule’ with out Marching band 4th period
• How many ‘Elective schedule’ with Marching band 4th period are possible?
Filling in the Forth Period piece is easy.
Third Period Class
choices
# choices
Forth Period Class
M
1 choices
Then we recall that the school regulations say if a student is in Marching band during
forth period, then they have to take Band or Choir third period, so we can fill in the
rest of the table:
choices
# choices
Third Period Class
B,D
2 choices
Forth Period Class
M
1 choices
So there are 2 × 1 = 2 ‘Elective schedule’ with Marching band 4th period.
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• How many ‘Elective schedule’ with out Marching band 4th period are possible?
Again we start by listing our options for 4th period.
Third Period Class
choices
# choices
Forth Period Class
S, P, Y W,
4 choices
Then we ask ourselves what classes the students who are NOT in marching band can
take third period:
choices
# choices
Third Period Class
B, D, G, C
4 choices
Forth Period Class
S, P, Y W,
4 choices
So there are 4 × 4 = 16 ‘Elective schedule’ with out Marching band 4th
period.
THUS, there are 2 + 16 = 18 different ‘Elective schedules’ possible.
Those 18 ’Elective Schedules’ are:
• (from Elective schedule’ with Marching band 4th period)
BM, CM
• (from Elective schedule’ with out Marching band 4th period)
BS, BP, BY, BW
DS, DP, DY, DW
GS, GP, GY, GW
CS, CP, CY, CW
Notice that BS stands for Band third period & Sculpture forth period, BP stands for
Band third period & Photography forth period, etc.
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