3.4C-Combinations
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•
•
•
•
.
Combinations of n objects taken r at a time
Order does NOT matter
nCr = n!/(n-r)!n!
n=# objects r=# taken
Calculator: n, MATH, → PRB, 3:nCr, r, ENTER
Ex: 16 construction companies are bidding to do
the job. How many different combinations of 4
companies can be chosen?
3.4C-Combinations
•
•
•
•
•
.
Combinations of n objects taken r at a time
Order does NOT matter
nCr = n!/(n-r)!n!
n=# objects r=# taken
Calculator: n, MATH, → PRB, 3:nCr, r, ENTER
Ex: 16 construction companies are bidding to do
the job. How many different combinations of 4
companies can be chosen?
no particular order
3.4C-Combinations
•
•
•
•
•
Combinations of n objects taken r at a time
Order does NOT matter
nCr = n!/(n-r)!n!
n=# objects r=# taken
Calculator: n, MATH, → PRB, 3:nCr, r, ENTER
Ex: 16 construction companies are bidding to do
the job. How many different combinations of 4
companies can be chosen?
no particular order
.
₁₆C₄ =
3.4C-Combinations
•
•
•
•
•
Combinations of n objects taken r at a time
Order does NOT matter
nCr = n!/(n-r)!n!
n=# objects r=# taken
Calculator: n, MATH, → PRB, 3:nCr, r, ENTER
Ex: 16 construction companies are bidding to do
the job. How many different combinations of 4
companies can be chosen?
no particular order
.
₁₆C₄ = 16!/(16-4)!4!
3.4C-Combinations
•
•
•
•
•
Combinations of n objects taken r at a time
Order does NOT matter
nCr = n!/(n-r)!n!
n=# objects r=# taken
Calculator: n, MATH, → PRB, 3:nCr, r, ENTER
Ex: 16 construction companies are bidding to do
the job. How many different combinations of 4
companies can be chosen?
no particular order
.
₁₆C₄ = 16!/(16-4)!4! = 16!/(12!4!)
3.4C-Combinations
•
•
•
•
•
Combinations of n objects taken r at a time
Order does NOT matter
nCr = n!/(n-r)!n!
n=# objects r=# taken
Calculator: n, MATH, → PRB, 3:nCr, r, ENTER
Ex: 16 construction companies are bidding to do
the job. How many different combinations of 4
companies can be chosen?
no particular order
.
₁₆C₄ = 16!/(16-4)!4! = 16!/(12!4!)=1820
More Examples:
• There are 16 employees. In how many ways can
an advisory committee of 3 be chosen?
• A word has 1 M, 4 I’s, 4 S’s & 2 P’s. What is the
probability they can be arranged to spell
MISSISSIPPI?
.
More Examples:
• There are 16 employees. In how many ways can
an advisory committee of 3 be chosen?
order NOT important
• A word has 1 M, 4 I’s, 4 S’s & 2 P’s. What is the
probability they can be arranged to spell
MISSISSIPPI?
.
More Examples:
• There are 16 employees. In how many ways can
an advisory committee of 3 be chosen?
order NOT important
₁₆C₃ =
• A word has 1 M, 4 I’s, 4 S’s & 2 P’s. What is the
probability they can be arranged to spell
MISSISSIPPI?
.
More Examples:
• There are 16 employees. In how many ways can
an advisory committee of 3 be chosen?
order NOT important
₁₆C₃ = 16!/(16-3)!3!
• A word has 1 M, 4 I’s, 4 S’s & 2 P’s. What is the
probability they can be arranged to spell
MISSISSIPPI?
.
More Examples:
• There are 16 employees. In how many ways can
an advisory committee of 3 be chosen?
order NOT important
₁₆C₃ = 16!/(16-3)!3! = 16!/(13!3!)
• A word has 1 M, 4 I’s, 4 S’s & 2 P’s. What is the
probability they can be arranged to spell
MISSISSIPPI?
.
More Examples:
• There are 16 employees. In how many ways can
an advisory committee of 3 be chosen?
order NOT important
₁₆C₃ = 16!/(16-3)!3! = 16!/(13!3!) = 560
• A word has 1 M, 4 I’s, 4 S’s & 2 P’s. What is the
probability they can be arranged to spell
MISSISSIPPI?
.
More Examples:
• There are 16 employees. In how many ways can
an advisory committee of 3 be chosen?
order NOT important
₁₆C₃ = 16!/(16-3)!3! = 16!/(13!3!) = 560
• A word has 1 M, 4 I’s, 4 S’s & 2 P’s. What is the
probability they can be arranged to spell
MISSISSIPPI?
order IS important (distinguishable permutation)
.
More Examples:
• There are 16 employees. In how many ways can
an advisory committee of 3 be chosen?
order NOT important
₁₆C₃ = 16!/(16-3)!3! = 16!/(13!3!) = 560
• A word has 1 M, 4 I’s, 4 S’s & 2 P’s. What is the
probability they can be arranged to spell
MISSISSIPPI?
order IS important (distinguishable permutation)
.
11!/(1!4!4!2!)
More Examples:
• There are 16 employees. In how many ways can
an advisory committee of 3 be chosen?
order NOT important
₁₆C₃ = 16!/(16-3)!3! = 16!/(13!3!) = 560
• A word has 1 M, 4 I’s, 4 S’s & 2 P’s. What is the
probability they can be arranged to spell
MISSISSIPPI?
order IS important (distinguishable permutation)
.
11!/(1!4!4!2!) = 34650 possible & 1 favorable
1/34650
More Examples:
• There are 16 employees. In how many ways can
an advisory committee of 3 be chosen?
order NOT important
₁₆C₃ = 16!/(16-3)!3! = 16!/(13!3!) = 560
• A word has 1 M, 4 I’s, 4 S’s & 2 P’s. What is the
probability they can be arranged to spell
MISSISSIPPI?
order IS important (distinguishable permutation)
.
11!/(1!4!4!2!) = 34650 possible & 1 favorable
1/34650 = .000029
More Examples
• A word has 1 L, 2 E’s, 2 T’s, & 1R. What is the
probability they will be arranged to spell LETER?
• Find the probability of being dealt 5 diamonds in
a row from a standard deck of 52.
.
More Examples
• A word has 1 L, 2 E’s, 2 T’s, & 1R. What is the
probability they will be arranged to spell LETER?
Order IS important
• Find the probability of being dealt 5 diamonds in
a row from a standard deck of 52.
.
More Examples
• A word has 1 L, 2 E’s, 2 T’s, & 1R. What is the
probability they will be arranged to spell LETER?
Order IS important Distinguishable permutation
• Find the probability of being dealt 5 diamonds in
a row from a standard deck of 52.
.
More Examples
• A word has 1 L, 2 E’s, 2 T’s, & 1R. What is the
probability they will be arranged to spell LETER?
Order IS important Distinguishable permutation
6!/(1!2!2!1!) =
• Find the probability of being dealt 5 diamonds in
a row from a standard deck of 52.
.
More Examples
• A word has 1 L, 2 E’s, 2 T’s, & 1R. What is the
probability they will be arranged to spell LETER?
Order IS important Distinguishable permutation
6!/(1!2!2!1!) = 180 possible, 1 desirable
• Find the probability of being dealt 5 diamonds in
a row from a standard deck of 52.
.
More Examples
• A word has 1 L, 2 E’s, 2 T’s, & 1R. What is the
probability they will be arranged to spell LETER?
Order IS important Distinguishable permutation
6!/(1!2!2!1!) = 180 possible, 1 desirable = 1/180
• Find the probability of being dealt 5 diamonds in
a row from a standard deck of 52.
.
More Examples
• A word has 1 L, 2 E’s, 2 T’s, & 1R. What is the
probability they will be arranged to spell LETER?
Order IS important Distinguishable permutation
6!/(1!2!2!1!) = 180 possible, 1 desirable = 1/180
= .0056
• Find the probability of being dealt 5 diamonds in
a row from a standard deck of 52.
.
More Examples
• A word has 1 L, 2 E’s, 2 T’s, & 1R. What is the
probability they will be arranged to spell LETER?
Order IS important Distinguishable permutation
6!/(1!2!2!1!) = 180 possible, 1 desirable = 1/180
= .0056
• Find the probability of being dealt 5 diamonds in
a row from a standard deck of 52.
Order NOT important
.
More Examples
• A word has 1 L, 2 E’s, 2 T’s, & 1R. What is the
probability they will be arranged to spell LETER?
Order IS important Distinguishable permutation
6!/(1!2!2!1!) = 180 possible, 1 desirable = 1/180
= .0056
• Find the probability of being dealt 5 diamonds in
a row from a standard deck of 52.
Order NOT important
P(desirable)/P(Total)
.
More Examples
• A word has 1 L, 2 E’s, 2 T’s, & 1R. What is the
probability they will be arranged to spell LETER?
Order IS important Distinguishable permutation
6!/(1!2!2!1!) = 180 possible, 1 desirable = 1/180
= .0056
• Find the probability of being dealt 5 diamonds in
a row from a standard deck of 52.
Order NOT important
P(desirable)/P(Total)=P(5 of 13 D)/P(5 of 52)
.
More Examples
• A word has 1 L, 2 E’s, 2 T’s, & 1R. What is the
probability they will be arranged to spell LETER?
Order IS important Distinguishable permutation
6!/(1!2!2!1!) = 180 possible, 1 desirable = 1/180
= .0056
• Find the probability of being dealt 5 diamonds in
a row from a standard deck of 52.
Order NOT important
P(desirable)/P(Total)=P(5 of 13 D)/P(5 of 52)
. ₁₃C₅/₅₂C₅
More Examples
• A word has 1 L, 2 E’s, 2 T’s, & 1R. What is the
probability they will be arranged to spell LETER?
Order IS important Distinguishable permutation
6!/(1!2!2!1!) = 180 possible, 1 desirable = 1/180
= .0056
• Find the probability of being dealt 5 diamonds in
a row from a standard deck of 52.
Order NOT important
P(desirable)/P(Total)=P(5 of 13 D)/P(5 of 52)
. ₁₃C₅/₅₂C₅ = 1287/2598960
More Examples
• A word has 1 L, 2 E’s, 2 T’s, & 1R. What is the
probability they will be arranged to spell LETER?
Order IS important Distinguishable permutation
6!/(1!2!2!1!) = 180 possible, 1 desirable = 1/180
= .0056
• Find the probability of being dealt 5 diamonds in
a row from a standard deck of 52.
Order NOT important
P(desirable)/P(Total)=P(5 of 13 D)/P(5 of 52)
. ₁₃C₅/₅₂C₅ = 1287/2598960 = .0005
More Examples:
• A group contains 5 men and 7 women. If 3 are
selected at random, what is the probability
that all 3 are men?
More Examples:
• A group contains 5 men and 7 women. If 3 are
selected at random, what is the probability
that all 3 are men?
Order is NOT important
More Examples:
• A group contains 5 men and 7 women. If 3 are
selected at random, what is the probability
that all 3 are men?
Order is NOT important
P(3 M)·P(0W)
More Examples:
• A group contains 5 men and 7 women. If 3 are
selected at random, what is the probability
that all 3 are men?
Order is NOT important
P(3 M)·P(0W) = ₅C₃· ₇C₀ =
More Examples:
• A group contains 5 men and 7 women. If 3 are
selected at random, what is the probability
that all 3 are men?
Order is NOT important
P(3 M)·P(0W) = ₅C₃· ₇C₀ = 10·1
More Examples:
• A group contains 5 men and 7 women. If 3 are
selected at random, what is the probability
that all 3 are men?
Order is NOT important
P(3 M)·P(0W) = ₅C₃· ₇C₀ = 10·1 =10
P(choose 3 from 12)
More Examples:
• A group contains 5 men and 7 women. If 3 are
selected at random, what is the probability
that all 3 are men?
Order is NOT important
P(3 M)·P(0W) = ₅C₃· ₇C₀ = 10·1 =10
P(choose 3 from 12) = ₁₂C₃ =
More Examples:
• A group contains 5 men and 7 women. If 3 are
selected at random, what is the probability
that all 3 are men?
Order is NOT important
P(3 M)·P(0W) = ₅C₃· ₇C₀ = 10·1 =10
P(choose 3 from 12) = ₁₂C₃ = 220
More Examples:
• A group contains 5 men and 7 women. If 3 are
selected at random, what is the probability
that all 3 are men?
Order is NOT important
P(3 M)·P(0W) = ₅C₃· ₇C₀ = 10·1 =10
P(choose 3 from 12) = ₁₂C₃ = 220
P(all men) = P(3 M)·P( 0 W) / P(3 from 12)
More Examples:
• A group contains 5 men and 7 women. If 3 are
selected at random, what is the probability
that all 3 are men?
Order is NOT important
P(3 M)·P(0W) = ₅C₃· ₇C₀ = 10·1 =10
P(choose 3 from 12) = ₁₂C₃ = 220
P(all men) = P(3 M)·P( 0 W) / P(3 from 12)
10/220 =
More Examples:
• A group contains 5 men and 7 women. If 3 are
selected at random, what is the probability
that all 3 are men?
Order is NOT important
P(3 M)·P(0W) = ₅C₃· ₇C₀ = 10·1 =10
P(choose 3 from 12) = ₁₂C₃ = 220
P(all men) = P(3 M)·P( 0 W) / P(3 from 12)
10/220 = .045