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Combinations with Restrictions>
Combinations with Restrictions 1Combinations with Restrictions 1
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Question 1 of 6
1. Question
How many ways can a soccer team of 11 players be selected from a squad of 13 existing players if the captain must be included?- (66)
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Use the combinations formula to find the number of ways an item can be chosen (r) from the total number of items (n).Remember that order is not important in Combinations.Combination Formula
nCr=n!(n−r)!r!First, find the ways that the captain can be selected. There is only one captain, which means:r=1n=1nCr = n!(n−r)!r! Combination Formula 1C1 = 1!(1−1)!1! Substitute the values of r and n = 1!0!1! = 1 0!=1 Keeping in mind that the captain has already been picked, find the different ways that the 10 other players (r) can be selected from a total of 12 players (n)r=10n=12nCr = n!(n−r)!r! Combination Formula 12C10 = 12!(12−10)!10! Substitute the values of r and n = 12!2!10! = 12⋅11⋅10⋅9⋅8⋅7⋅6⋅5⋅4⋅3⋅2⋅12⋅1⋅10⋅9⋅8⋅7⋅6⋅5⋅4⋅3⋅2⋅1 = 1322 Cancel like terms = 66 Finally, multiply the two solved combinationsNumber of ways the captain can be selected=1Number of ways the other players can be selected=661⋅66 = 66 Therefore, there are 66 ways of selecting a soccer team of 11 players if the captain must be included.66 -
Question 2 of 6
2. Question
How many ways can we form a 5 man basketball team from a group of 9 players if the captain and the vice captain must be included?- (35)
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Use the combinations formula to find the number of ways an item can be chosen (r) from the total number of items (n).Remember that order is not important in Combinations.Combination Formula
nCr=n!(n−r)!r!First, find the ways that the captain and the vice captain can be selected. There is only one captain and one vice captain, which means:r=2n=2nCr = n!(n−r)!r! Combination Formula 2C2 = 2!(2−2)!2! Substitute the values of r and n = 2!0!2! = 2⋅12⋅1 0!=1 = 1 Keeping in mind that the captain and the vice captain have already been picked, find the different ways that the 3 other players (r) can be selected from a total of 7 players (n)r=3n=7nCr = n!(n−r)!r! Combination Formula 7C3 = 7!(7−3)!3! Substitute the values of r and n = 7!4!3! = 7⋅6⋅5⋅4⋅3⋅2⋅14⋅3⋅2⋅1⋅3⋅2⋅1 = 7⋅6⋅53⋅2⋅1 Cancel like terms = 7⋅51 3×2=6 = 35 Finally, multiply the two solved combinationsNumber of ways the captain and the vice captain can be selected=1Number of ways the other players can be selected=351⋅35 = 35 Therefore, there are 35 ways of selecting a basketball team of 5 players if the captain and the vice captain must be included.35 -
Question 3 of 6
3. Question
How many ways can students answer 6 items on a 10 items quiz if they are required to answer the first two questions?- (70)
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Use the combinations formula to find the number of ways an item can be chosen (r) from the total number of items (n).Remember that order is not important in Combinations.Combination Formula
nCr=n!(n−r)!r!First, find the ways that the two questions can be answered. There are only two required questions to answer, which means:r=2n=2nCr = n!(n−r)!r! Combination Formula 2C2 = 2!(2−2)!2! Substitute the values of r and n = 2!0!2! = 2⋅12⋅1 0!=1 = 1 Keeping in mind that the students already answered two questions, find the different ways that the 4 other questions (r) can be answered from a total of 8 remaining questions (n)r=4n=8nCr = n!(n−r)!r! Combination Formula 8C4 = 8!(8−4)!4! Substitute the values of r and n = 8!4!4! = 6⋅7⋅6⋅5⋅4⋅3⋅2⋅14⋅3⋅2⋅1⋅4⋅3⋅2⋅1 = 8⋅7⋅6⋅54⋅3⋅2⋅1 Cancel like terms = 7⋅6⋅53⋅1 4×2=8 = 2103 4×2=8 = 70 Finally, multiply the two solved combinationsNumber of ways the first two questions can be answered=1Number of ways other questions can be answered=701⋅70 = 70 Therefore, there are 70 ways of answering 6 items on a 10 items quiz if the students are required to answer the first two questions.70 -
Question 4 of 6
4. Question
How many ways can a 12-member panel be formed from a total pool of 38 people, if 2 of them are certain to be part of that panel?- 1.
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Use the combinations formula to find the number of ways an item can be chosen (r) from the total number of items (n).Remember that order is not important in Combinations.Combination Formula
nCr=n!(n−r)!r!First, find the ways that the 2 people who are certain to be part of the panel can be arranged. There are only 2 positions available for them, which means:r=2n=2nCr = n!(n−r)!r! Combination Formula 2C2 = 2!(2−2)!2! Substitute the values of r and n = 2!0!2! = 2⋅12⋅1 0!=1 = 1 Keeping in mind that two people are already part of the panel, find the different ways that the 10 other members (r) can be chosen from a total of 36 remaining people (n)r=10n=36nCr = n!(n−r)!r! Combination Formula 36C10 = 36!(36−10)!10! Substitute the values of r and n = 36!26!10! = 254 186 856 Use the calculator’s factorial function for large numbers Finally, multiply the two solved combinationsNumber of ways the two people are arranged=1Number of ways other people can be chosen=254 186 8561⋅254 186 856 = 254 186 856 Therefore, there are 254 186 856 ways of forming a 12-member panel from a total pool of 38 people, if 2 of them are certain to be part of the panel.254 186 856 -
Question 5 of 6
5. Question
In how many ways can you be dealt 4 Kings and 1 other card using a standard 52-card deck?- (48)
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Use the combinations formula to find the number of ways an item can be chosen (r) from the total number of items (n).Remember that order is not important in Combinations.Combination Formula
nCr=n!(n−r)!r!First, find the ways that the 4 King cards can be dealt. There are only 4 King cards available, which means:r=4n=4nCr = n!(n−r)!r! Combination Formula 4C4 = 4!(4−4)!4! Substitute the values of r and n = 4!0!4! = 4⋅3⋅2⋅14⋅3⋅2⋅1 0!=1 = 1 Keeping in mind that 4 cards have already been dealt, find the different ways that 1 other card (r) can be chosen from a total of 48 remaining cards (n)r=1n=48nCr = n!(n−r)!r! Combination Formula 48C1 = 48!(48−1)!1! Substitute the values of r and n = 48!47!1! = 48⋅47!47! = 48 47!47! cancels out Finally, multiply the two solved combinationsNumber of ways the 4 King cards can be dealt=1Number of ways one other card can be dealt=481⋅48 = 48 Therefore, there are 48 ways of dealing 4 Kings and one other card from a standard deck of 52 cards.48 -
Question 6 of 6
6. Question
In how many ways can you be dealt 3 Queens and 2 other cards using a standard 52-card deck?- (4512)
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Chapters- Chapters
Use the combinations formula to find the number of ways an item can be chosen (r) from the total number of items (n).Remember that order is not important in Combinations.Combination Formula
nCr=n!(n−r)!r!First, find the ways that the 3 Queens can be dealt. There are 4 Queens available, which means:r=3n=4nCr = n!(n−r)!r! Combination Formula 4C3 = 4!(4−3)!3! Substitute the values of r and n = 4!1!3! = 4⋅3⋅2⋅13⋅2⋅1 0!=1 = 4 Keeping in mind that the 4 Queens cannot be chosen again, find the different ways that 2 other cards (r) can be chosen from a total of 48 remaining cards (n)r=2n=48nCr = n!(n−r)!r! Combination Formula 48C2 = 48!(48−2)!2! Substitute the values of r and n = 48!46!2! = 48⋅47⋅46!46!⋅2⋅1 = 22562 46!46! cancels out = 1128 Finally, multiply the two solved combinationsNumber of ways the 3 Queens can be dealt=4Number of ways two other cards can be dealt=11284⋅1128 = 4512 Therefore, there are 4512 ways of dealing 3 Queens and two other cards from a standard deck of 52 cards.4512
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