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Permutations with Restrictions 4Permutations with Restrictions 4
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Question 1 of 7
1. Question
Find the number of ways 3 unique cats and 5 unique dogs can be seated in a straight line, given that the 3 cats must always sit together- (4320)
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Use the permutations formula to find the number of ways an item can be arranged (r) from the total number of items (n).Remember that order is important in Permutations.Permutation Formula if (n=r)
nPn=n!Solve the number of arrangements if the three cats are treated as one entity and the number of arrangements for those 3 cats, then multiply them.First, treat the 3 cats as one. This leaves us with 6 animals (r) to be seated in 6 places in the straight line (n)n=r=4nPn = n! Permutation Formula (if n=r) 6P6 = 6! Substitute the value of n = 6⋅5⋅4⋅3⋅2⋅1 = 720 There are 720 ways to arrange 6 animals.Next, arrange the 3 cats (r) in 3 positions (n)n=r=3nPn = n! Permutation Formula (if n=r) 3P3 = 3! Substitute the value of n = 3⋅2⋅1 = 6 There are 6 ways to arrange three catsFinally, multiply the two solved permutations6P6=7203P3=6720⋅6 = 4320 Therefore, there are 4320 ways of arranging 3 cats and 5 dogs if the cats should be seated together4320 -
Question 2 of 7
2. Question
A bakery has a section for Muffins, Donuts and Cookies. How many ways can 6 muffins, 3 donuts, and 2 cookies be arranged if they must stay in their respective sections?- 1.
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Use the permutations formula to find the number of ways an item can be arranged (r) from the total number of items (n).Remember that order is important in Permutations.Permutation Formula if (n=r)
nPn=n!Solve and multiply four permutations for: the sections, muffins, donuts and cookies.First, arrange 3 sections (r) in 3 positions (n)n=r=3nPn = n! Permutation Formula (if n=r) 3P3 = 3! Substitute the value of n = 3⋅2⋅1 = 6 There are 6 ways to arrange 3 sections.Next, arrange 6 muffins (r) in 6 positions (n)n=r=6nPn = n! Permutation Formula (if n=r) 6P6 = 6! Substitute the value of n = 6⋅5⋅4⋅3⋅2⋅1 = 720 There are 720 ways to arrange 6 muffinsThen, arrange 3 donuts (r) in 3 positions (n)n=r=3nPn = n! Permutation Formula (if n=r) 3P3 = 3! Substitute the value of n = 3⋅2⋅1 = 6 There are 6 ways to arrange 3 donutsNow, arrange 2 cookies (r) in 2 positions (n)n=r=2nPn = n! Permutation Formula (if n=r) 2P2 = 3! Substitute the value of n = 2⋅1 = 2 There are 2 ways to arrange 2 cookiesFinally, multiply the four solved permutationssections=6muffins=720donuts=6cookies=26⋅720⋅6⋅2 = 51 840 Therefore, there are 51 840 ways of arranging 6 muffins, 3 donuts and 2 cookies in their respective sections51 840 -
Question 3 of 7
3. Question
A music playlist has a section for Rock, Jazz and Pop music. How many ways can 4 rock songs, 3 jazz songs, and 2 pop songs be arranged if they must stay in their respective genres?- (1728)
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Use the permutations formula to find the number of ways an item can be arranged (r) from the total number of items (n).Remember that order is important in Permutations.Permutation Formula if (n=r)
nPn=n!Solve and multiply four permutations for: the genres, rock songs, jazz songs and pop songs.First, arrange 3 genres (r) in 3 positions (n)n=r=3nPn = n! Permutation Formula (if n=r) 3P3 = 3! Substitute the value of n = 3⋅2⋅1 = 6 There are 6 ways to arrange 3 genres.Next, arrange 4 rock songs (r) in 4 positions (n)n=r=4nPn = n! Permutation Formula (if n=r) 4P4 = 4! Substitute the value of n = 4⋅3⋅2⋅1 = 24 There are 24 ways to arrange 4 rock songsThen, arrange 3 jazz songs (r) in 3 positions (n)n=r=3nPn = n! Permutation Formula (if n=r) 3P3 = 3! Substitute the value of n = 3⋅2⋅1 = 6 There are 6 ways to arrange 3 jazz songsNow, arrange 2 pop songs (r) in 2 positions (n)n=r=2nPn = n! Permutation Formula (if n=r) 2P2 = 3! Substitute the value of n = 2⋅1 = 2 There are 2 ways to arrange 2 pop songsFinally, multiply the four solved permutationsgenres=6rock songs=24jazz songs=6pop songs=26⋅24⋅6⋅2 = 1728 Therefore, there are 1728 ways of arranging 4 rock songs, 3 jazz songs and 2 pop songs in their respective sections1728 -
Question 4 of 7
4. Question
A music playlist has a section for Rock, Jazz and Pop music. Given that the rock genre must be placed last, how many ways can 4 rock songs, 3 jazz songs, and 2 pop songs be arranged?- (576)
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Use the permutations formula to find the number of ways an item can be arranged (r) from the total number of items (n).Remember that order is important in Permutations.Permutation Formula if (n=r)
nPn=n!Solve and multiply four permutations for: the genres, rock songs, jazz songs and pop songs.First, remember that the rock genre must stay last. This means we are left to arrange 2 genres (r) in 2 positions (n)n=r=2nPn = n! Permutation Formula (if n=r) 2P2 = 2! Substitute the value of n = 2⋅1 = 2 There are 2 ways to arrange 3 genres if rock must stay last.Next, arrange 4 rock songs (r) in 4 positions (n)n=r=4nPn = n! Permutation Formula (if n=r) 4P4 = 4! Substitute the value of n = 4⋅3⋅2⋅1 = 24 There are 24 ways to arrange 4 rock songsThen, arrange 3 jazz songs (r) in 3 positions (n)n=r=3nPn = n! Permutation Formula (if n=r) 3P3 = 3! Substitute the value of n = 3⋅2⋅1 = 6 There are 6 ways to arrange 3 jazz songsNow, arrange 2 pop songs (r) in 2 positions (n)n=r=2nPn = n! Permutation Formula (if n=r) 2P2 = 3! Substitute the value of n = 2⋅1 = 2 There are 2 ways to arrange 2 pop songsFinally, multiply the four solved permutationsgenres=2rock songs=24jazz songs=6pop songs=22⋅24⋅6⋅2 = 576 Therefore, there are 576 ways of arranging 4 rock songs, 3 jazz songs and 2 pop songs if the rock genre must stay last576 -
Question 5 of 7
5. Question
A pizza booth has 6 seats — 3 on the left side and 3 on the right. How many ways can six people be seated in the booth if 2 girls insist that they sit on the right side?- (144)
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Use the permutations formula to find the number of ways an item can be arranged (r) from the total number of items (n).Remember that order is important in Permutations.Permutation Formula
nPr=n!(n−r)!Solve the permutation for the two girls who want to sit on the right side and the permutation for the remaining 4 people, then multiply.First, arrange 2 girls (r) in the 3 seats on the right side (n)n=3r=2nPr = n!(n−r)! Permutation Formula 3P2 = 3!(3−2)! Substitute the value of n and r = 3!1! = 3⋅2⋅11 = 6 Cancel like terms and evaluate There are 6 ways to arrange 2 girls in the three seats on the right.Next, arrange 4 people (r) in 4 seats (n)n=r=4nPn = n! Permutation Formula (if n=r) 4P4 = 4! Substitute the value of n = 4⋅3⋅2⋅1 = 24 There are 24 ways to arrange 4 peopleFinally, multiply the two solved permutations2 girls=64 people=246⋅24 = 144 Therefore, there are 144 ways of arranging 6 people in the pizza booth if 2 girls want to sit on the right side144 -
Question 6 of 7
6. Question
A learjet has 8 seats — 4 on the left side and 4 on the right. How many ways can eight people be seated in the learjet if 3 passengers insist to be on the right side and 2 passengers insist to be on the left side?- (1728)
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Use the permutations formula to find the number of ways an item can be arranged (r) from the total number of items (n).Remember that order is important in Permutations.Permutation Formula
nPr=n!(n−r)!Solve the permutation for the 3 passengers who want to sit on the right side, the 2 passengers who want to sit on the left side and the permutation for the remaining 3 people, then multiply.First, arrange 3 passengers (r) in the 4 seats on the right side (n)n=4r=3nPr = n!(n−r)! Permutation Formula 4P3 = 4!(4−3)! Substitute the value of n and r = 4!1! = 4⋅3⋅2⋅11 = 24 Cancel like terms and evaluate There are 24 ways to arrange 3 passengers in the four seats on the right.Next, arrange 2 passengers (r) in the 4 seats on the left side (n)n=4r=2nPr = n!(n−r)! Permutation Formula 4P2 = 4!(4−2)! Substitute the value of n and r = 4!2! = 4⋅3⋅2⋅12⋅1 = 12 Cancel like terms and evaluate There are 12 ways to arrange 2 passengers in the four seats on the left.Now, arrange 3 remaining passengers (r) in 3 remaining seats (n)n=r=3nPn = n! Permutation Formula (if n=r) 3P3 = 3! Substitute the value of n = 3⋅2⋅1 = 6 There are 6 ways to arrange 3 passengersFinally, multiply the three solved permutationspassengers who want to sit on the right=24passengers who want to sit on the left=12other passengers=624⋅12⋅6 = 1728 Therefore, there are 1728 ways of arranging 8 passengers in the learjet if 3 want to sit on the right and 2 want to sit on the left.1728 -
Question 7 of 7
7. Question
A speedboat has 4 seats — 2 on the port side (left side) and 2 on the right. One girl wishes to sit on the port side. If there are 4 passengers on the boat, what is the probability of that happening?Write fractions in the format “a/b”- (1/2)
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Permutation Formula
nPr=n!(n−r)!Probability
favourableoutcometotaloutcomeSolve the the number of ways the girl could get a seat on the port side and divide it by the total number of ways all 4 passengers can be arranged.First, place 1 girl (r) in the 2 seats on the port side (n)n=2r=1nPr = n!(n−r)! Permutation Formula 2P1 = 2!(2−1)! Substitute the value of n and r = 2!1! = 2⋅11 = 2 Cancel like terms There are 2 ways to place 1 girl on the port side.Now, arrange 3 remaining passengers (r) in 3 remaining seats (n)n=r=3nPn = n! Permutation Formula (if n=r) 3P3 = 3! Substitute the value of n = 3⋅2⋅1 = 6 There are 6 ways to arrange 3 passengersMultiply the two solved permutationsgirl who wants to sit on the port side=2other passengers=62⋅6 = 12 Hence, there are 12 ways of arranging 4 passengers on the boat if a girl wants to sit on the port side. This is the favourable outcomeNext, find the total number of ways to arrange all 4 passengers (r) in the 4 seats (n)n=r=4nPn = n! Permutation Formula (if n=r) 4P4 = 4! Substitute the value of n = 4⋅3⋅2⋅1 = 24 Hence, there are 24 ways to arrange 4 passengers. This is the total outcomeFinally, solve for the probabilityProbability = favourableoutcometotaloutcome = 1224 = 12 The probability that the girl gets to sit on the port side is 1212
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