matura.mkSecondary school › Counting and probability
Permutations
Permutations of n and of r elements, grouped and identical elements, paths on a grid, circular permutations. 10 problems with step-by-step solutions.
Solve each problem in your notebook first, then open the solution.
Problem 1 Permutations of n elements
a) In how many ways can 6 different books stand on a shelf?
b) In how many ways can 4 runners finish a race (no ties)?
c) 4 buses wait to leave and 2 buses wait to arrive at a station with one gate. In how many orders can the 6 operations be done?
Show solution
a) \(6! = 720\).
b) \(4! = 24\).
c) \(6! = 720\).
Problem 2 Permutations of r elements
\(P(n, r) = \dfrac{n!}{(n - r)!}\).
a) Calculate \(P(6, 2)\cdot P(5, 3)\), \(P(8, 3)\), \(P(n, 1)\).
b) \(P(n, 2) = 42\). Find \(n\). c) \(P(n, 3) = 6\cdot P(n, 2)\). Find \(n\).
Show solution
a) \(30\cdot 60 = 1800\); \(8\cdot 7\cdot 6 = 336\); \(n\).
b) \(n(n - 1) = 42\), \(n = 7\).
c) \(n(n - 1)(n - 2) = 6n(n - 1)\), \(n - 2 = 6\), \(n = 8\).
Problem 3 Ordered selections
a) From a club of 15 members we choose a president, a secretary and a treasurer. In how many ways?
b) How many three-letter "words" with different letters can be formed from the letters of the word PLANET?
c) How many words of at least 4 letters, with no repeated letter, can be formed from the letters of the word NUMBER?
Show solution
a) \(P(15, 3) = 15\cdot 14\cdot 13 = 2730\).
b) \(P(6, 3) = 120\).
c) \(P(6, 4) + P(6, 5) + P(6, 6) = 360 + 720 + 720 = 1800\).
Problem 4 Shelves and groups
a) Ana has 6 math, 5 history and 4 art books (all different). She puts 2 math books on the first shelf, 3 history books on the second and 4 art books on the third. In how many ways?
b) From the digits \(\{1, 2, 4, 5, 6, 8, 9\}\) three-digit numbers with different digits are formed. How many do not contain 4? How many contain 4? How many contain 1 or 9?
Show solution
a) \(P(6, 2)\cdot P(5, 3)\cdot P(4, 4) = 30\cdot 60\cdot 24 = 43\,200\).
b) All: \(P(7, 3) = 210\). Without 4: \(P(6, 3) = 120\). With 4: \(210 - 120 = 90\). Without 1 and without 9: \(P(5, 3) = 60\), so with 1 or 9: \(210 - 60 = 150\).
Problem 5 Grouped elements
a) How many arrangements of all the letters of SOLVER have O and E next to each other?
b) In how many ways can 7 people sit in a row if two of them, Ivo and Mila, must not sit together?
c) 4 math, 3 physics and 2 chemistry books (all different) stand on a shelf. In how many ways if books of each subject stay together?
Show solution
a) Glue O and E: 5 blocks, \(5!\), and the pair can be OE or EO: \(2\cdot 120 = 240\).
b) All: \(7! = 5040\). Together: \(2\cdot 6! = 1440\). Not together: \(3600\).
c) \(3!\cdot 4!\cdot 3!\cdot 2! = 6\cdot 24\cdot 6\cdot 2 = 1728\).
Problem 6 Families in a row
a) A couple, a family of 4 and a family of 3 buy 9 tickets in one row. In how many ways can they sit if each family sits together?
b) Pupils from three classes (2, 4 and 3 pupils) walk in a line behind their teacher, and pupils of the same class walk together. How many orders are possible?
Show solution
a) \(3!\cdot 2!\cdot 4!\cdot 3! = 6\cdot 2\cdot 24\cdot 6 = 1728\).
b) The teacher is first. \(3!\cdot 2!\cdot 4!\cdot 3! = 1728\).
Problem 7 Identical elements
\(\dfrac{n!}{n_1!\,n_2!\cdots n_k!}\)
a) How many different numbers can be made from all the digits of 3 3 5 5 5 7?
b) How many distinguishable arrangements are there of the letters of STATISTICS?
c) 4 identical red, 3 identical blue and 2 identical green flags hang in a row. In how many ways?
d) A coin is tossed 8 times. In how many ways can we get exactly 5 heads?
Show solution
a) \(\frac{6!}{2!\,3!} = 60\).
b) S×3, T×3, I×2, A, C: \(\frac{10!}{3!\,3!\,2!} = 50\,400\).
c) \(\frac{9!}{4!\,3!\,2!} = 1260\).
d) \(\frac{8!}{5!\,3!} = 56\).
Problem 8 Paths on a grid
a) Streets form a grid. Marko walks from his house to school going only east or north; the school is 5 blocks east and 3 blocks north. How many shortest routes are there?
b) How many of these routes pass through the corner 2 blocks east and 1 block north of the house?
Show solution
a) A route is an arrangement of 5 E and 3 N: \(\frac{8!}{5!\,3!} = 56\).
b) To the corner: \(\frac{3!}{2!\,1!} = 3\); from the corner: 3 E and 2 N, \(\frac{5!}{3!\,2!} = 10\). Total \(30\).
Problem 9 Circular permutations
\(n\) different objects around a round table: \((n - 1)!\).
a) In how many ways can 7 friends sit at a round table?
b) A family of 2 parents and 5 children sits at a round table and the parents sit together. In how many ways?
c) 3 Italian, 4 French and 2 Greek guests sit at a round table and guests from the same country sit together. In how many ways?
d) In how many ways can 4 girls and 4 boys sit at a round table with 8 chairs so that no two girls sit together?
Show solution
a) \(6! = 720\).
b) Glue the parents: 6 objects around the table, \(5!\cdot 2 = 240\).
c) 3 groups around the table: \(2!\), inside: \(3!\,4!\,2!\). Total \(2\cdot 6\cdot 24\cdot 2 = 576\).
d) Seat the boys: \(3! = 6\). The girls go into the 4 gaps: \(4! = 24\). Total \(144\).
Problem 10 Necklaces and tables
a) In how many ways can 5 different beads be put on a necklace (which can be turned over)?
b) A room has a round table with 4 chairs and a bench for 3 people. In how many ways can 7 students sit?
Show solution
a) Circular: \(4! = 24\); turning over gives the same necklace: \(\frac{24}{2} = 12\).
b) Choose who sits at the table: \(C(7, 4) = 35\); arrange them: \(3! = 6\); the other 3 on the bench: \(3! = 6\). Total \(35\cdot 6\cdot 6 = 1260\).