Secondary school › Counting and probability

Counting principles

Addition and multiplication principle, numbers from digits and factorial. 4 problems with step-by-step solutions.

Solve each problem in your notebook first, then open the solution.

Problem 1 Addition and multiplication principle

a) A café offers 4 kinds of sandwiches and 3 kinds of salad. You take one sandwich or one salad. How many choices are there?

b) You take one sandwich and one salad and one of 5 drinks. How many different meals are there?

c) There are 3 roads from town A to B and 4 roads from B to C. How many ways are there to go from A to C and back, if you do not use the same road twice?

Show solution

a) Addition principle: \(4 + 3 = 7\).

b) Multiplication principle: \(4\cdot 3\cdot 5 = 60\).

c) There: \(3\cdot 4 = 12\). Back without repeating a road: \(3\cdot 2 = 6\). Total \(12\cdot 6 = 72\).

Problem 2 Numbers from digits

Using the digits \(\{0, 2, 3, 5, 7, 8\}\):

a) How many three-digit numbers can be formed (digits may repeat)?

b) How many three-digit numbers with different digits?

c) How many even three-digit numbers with different digits?

d) How many three-digit numbers greater than 500 with different digits?

Show solution

a) First digit not 0: \(5\cdot 6\cdot 6 = 180\).

b) \(5\cdot 5\cdot 4 = 100\).

c) Last digit 0: \(5\cdot 4 = 20\). Last digit 2 or 8: first digit 4 choices (not 0, not the last), middle 4: \(2\cdot 4\cdot 4 = 32\). Total \(52\).

d) First digit 5, 7 or 8: \(3\cdot 5\cdot 4 = 60\).

Problem 3 Codes and plates

a) A code has 2 letters (from 26) followed by 3 digits. How many codes are there? How many if no letter and no digit repeats?

b) A test has 10 true/false questions. In how many ways can it be answered?

Show solution

a) \(26^2\cdot 10^3 = 676\,000\). Without repetition: \(26\cdot 25\cdot 10\cdot 9\cdot 8 = 468\,000\).

b) \(2^{10} = 1024\).

Problem 4 Factorial

a) Calculate \(\dfrac{9!}{6!}\), \(\dfrac{12!}{10!\cdot 2!}\) and \(\dfrac{8! - 7!}{6!}\).

b) Simplify \(\dfrac{(n + 2)!}{n!}\) and \(\dfrac{(n - 49)!\,(n + 50)!}{(n - 50)!\,(n + 49)!}\).

c) Solve \(\dfrac{(n + 1)!}{(n - 1)!} = 56\).

Show solution

a) \(9\cdot 8\cdot 7 = 504\); \(\frac{12\cdot 11}{2} = 66\); \(\frac{7!(8 - 1)}{6!} = 7\cdot 7 = 49\).

b) \((n + 2)(n + 1)\); \((n - 49)(n + 50)\).

c) \((n + 1)n = 56\), \(n = 7\).