Combinations 1
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Question 1 of 6
1. Question
Evaluate4C34C3- (4)
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Use the combinations formula to find the number of ways an item can be chosen (r)(r) from the total number of items (n)(n).Remember that order is not important in Combinations.Combination Formula
nCr=n!(n−r)!r!nCr=n!(n−r)!r!Substitute the values to the combination formula.r=3r=3n=4n=4nCrnCr == n!(n−r)!r!n!(n−r)!r! Combination Formula 4C34C3 == 4!(4−3)!3!4!(4−3)!3! Substitute the values of rr and nn == 4!1!3!4!1!3! == 4⋅3⋅2⋅11⋅3⋅2⋅14⋅3⋅2⋅11⋅3⋅2⋅1 == 4141 Cancel like terms == 44 44 -
Question 2 of 6
2. Question
Evaluate10C410C4- (210)
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Use the combinations formula to find the number of ways an item can be chosen (r)(r) from the total number of items (n)(n).Remember that order is not important in Combinations.Combination Formula
nCr=n!(n−r)!r!nCr=n!(n−r)!r!Substitute the values to the combination formula.r=4r=4n=10n=10nCrnCr == n!(n−r)!r!n!(n−r)!r! Combination Formula 4C34C3 == 10!(10−4)!4!10!(10−4)!4! Substitute the values of rr and nn == 10!6!4!10!6!4! == 10⋅9⋅8⋅7⋅6⋅5⋅4⋅3⋅2⋅16⋅5⋅4⋅3⋅2⋅1⋅4⋅3⋅2⋅110⋅9⋅8⋅7⋅6⋅5⋅4⋅3⋅2⋅16⋅5⋅4⋅3⋅2⋅1⋅4⋅3⋅2⋅1 == 10⋅9⋅8⋅74⋅3⋅2⋅110⋅9⋅8⋅74⋅3⋅2⋅1 Cancel like terms == 10⋅9⋅73⋅110⋅9⋅73⋅1 4×2=84×2=8 == 63036303 == 210210 210210 -
Question 3 of 6
3. Question
Evaluate9C29C2- (36)
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Chapters- Chapters
Use the combinations formula to find the number of ways an item can be chosen (r)(r) from the total number of items (n)(n).Remember that order is not important in Combinations.Combination Formula
nCr=n!(n−r)!r!nCr=n!(n−r)!r!Substitute the values to the combination formula.r=2r=2n=9n=9nCrnCr == n!(n−r)!r!n!(n−r)!r! Combination Formula 4C34C3 == 9!(9−2)!2!9!(9−2)!2! Substitute the values of rr and nn == 9!7!2!9!7!2! == 9⋅8⋅7⋅6⋅5⋅4⋅3⋅2⋅17⋅6⋅5⋅4⋅3⋅2⋅1⋅2⋅19⋅8⋅7⋅6⋅5⋅4⋅3⋅2⋅17⋅6⋅5⋅4⋅3⋅2⋅1⋅2⋅1 == 9⋅82⋅19⋅82⋅1 Cancel like terms == 722722 == 3636 3636 -
Question 4 of 6
4. Question
How many ways can we pick 55 paintings from a gallery with 1414 paintings?- (2002)
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Chapters- Chapters
Use the combinations formula to find the number of ways an item can be chosen (r)(r) from the total number of items (n)(n).Remember that order is not important in Combinations.Combination Formula
nCr=n!(n−r)!r!nCr=n!(n−r)!r!We need to find the different ways that 55 paintings (r)(r) can be selected from a total of 1414 paintings (n)(n)r=5r=5n=14n=14nCrnCr == n!(n−r)!r!n!(n−r)!r! Combination Formula 13C1113C11 == 14!(14−5)!5!14!(14−5)!5! Substitute the values of rr and nn == 14!9!5!14!9!5! == 14⋅13⋅12⋅11⋅10⋅9⋅8⋅7⋅6⋅5⋅4⋅3⋅2⋅19⋅8⋅7⋅6⋅5⋅4⋅3⋅2⋅1⋅5⋅4⋅3⋅2⋅114⋅13⋅12⋅11⋅10⋅9⋅8⋅7⋅6⋅5⋅4⋅3⋅2⋅19⋅8⋅7⋅6⋅5⋅4⋅3⋅2⋅1⋅5⋅4⋅3⋅2⋅1 == 240240120240240120 Cancel like terms == 20022002 20022002 -
Question 5 of 6
5. Question
How many ways can a soccer team of 1111 players be selected from a squad of 1313 existing players?- 1.
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Chapters- Chapters
Use the combinations formula to find the number of ways an item can be chosen (r)(r) from the total number of items (n)(n).Remember that order is not important in Combinations.Combination Formula
nCr=n!(n−r)!r!nCr=n!(n−r)!r!We need to find the different ways that a team of 1111 players (r)(r) can be selected from a total of 1313 players (n)(n)r=11r=11n=13n=13nCrnCr == n!(n−r)!r!n!(n−r)!r! Combination Formula 13C1113C11 == 13!(13−11)!11!13!(13−11)!11! Substitute the values of rr and nn == 13!2!11!13!2!11! == 13⋅12⋅11⋅10⋅9⋅8⋅7⋅6⋅5⋅4⋅3⋅2⋅12⋅1⋅11⋅10⋅9⋅8⋅7⋅6⋅5⋅4⋅3⋅2⋅113⋅12⋅11⋅10⋅9⋅8⋅7⋅6⋅5⋅4⋅3⋅2⋅12⋅1⋅11⋅10⋅9⋅8⋅7⋅6⋅5⋅4⋅3⋅2⋅1 == 15621562 Cancel like terms == 7878 7878 -
Question 6 of 6
6. Question
How many ways can we form a 55 man basketball team from a group of 99 players?- (126)
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Chapters- Chapters
Use the combinations formula to find the number of ways an item can be chosen (r)(r) from the total number of items (n)(n).Remember that order is not important in Combinations.Combination Formula
nCr=n!(n−r)!r!nCr=n!(n−r)!r!We need to find the different ways that a team of 55 players (r)(r) can be selected from a total of 99 players (n)(n)r=5r=5n=9n=9nCrnCr == n!(n−r)!r!n!(n−r)!r! Combination Formula 9C59C5 == 9!(9−5)!5!9!(9−5)!5! Substitute the values of rr and nn == 9!4!5!9!4!5! == 9⋅8⋅7⋅6⋅5⋅4⋅3⋅2⋅14⋅3⋅2⋅1⋅5⋅4⋅3⋅2⋅19⋅8⋅7⋅6⋅5⋅4⋅3⋅2⋅14⋅3⋅2⋅1⋅5⋅4⋅3⋅2⋅1 = 9⋅8⋅7⋅64⋅3⋅2⋅1 Cancel like terms = 9⋅7⋅63⋅1 4×2=8 = 3783 = 126 126
Quizzes
- Factorial Notation
- Fundamental Counting Principle 1
- Fundamental Counting Principle 2
- Fundamental Counting Principle 3
- Combinations 1
- Combinations 2
- Combinations with Restrictions 1
- Combinations with Restrictions 2
- Combinations with Probability
- Basic Permutations 1
- Basic Permutations 2
- Basic Permutations 3
- Permutation Problems 1
- Permutation Problems 2
- Permutations with Repetitions 1
- Permutations with Repetitions 2
- Permutations with Restrictions 1
- Permutations with Restrictions 2
- Permutations with Restrictions 3
- Permutations with Restrictions 4