Lessons 28–29 · 2 lessons · M. A. Mirzaahmedov, Sh. N. Ismailov, A. Q. Amanov (algebra and analysis), B. Q. Haydarov (geometry). Mathematics Grade 11, Parts 1 and 2, 1st edition. ZAMIN NASHR, Tashkent, 2018
28
Combinatorics problems
Textbook, Part 2: pp. 27–32
GoalLearn to solve counting problems with the addition and multiplication rules, arrangements and permutations.
The main question of combinatorics is “in how many ways?”. When counting options we list them in order, so none is missed or repeated. Addition rule: if A can be chosen in m ways or B in another n ways, then “A or B” can be chosen in m + n ways. Multiplication rule: if A can be chosen in m ways and then B in n ways, “A and B” can be done in mn ways; for consecutive steps the numbers of ways are multiplied. The number of arrangements of k out of n elements without repetition is Aₙᵏ = n(n − 1)…(n − k + 1); with repetition it is n^k. The number of ways to order all n elements (permutations) is n! = 1 · 2 · … · n, with 0! = 1. An n-element set has 2ⁿ subsets, since each element is either in the subset or not.
Worked examples
A school canteen has 4 soups, 3 main courses and 2 drinks. A lunch can be assembled in 4 · 3 · 2 = 24 ways. Three-digit codes from the digits 1 to 6: without repetition 6 · 5 · 4 = 120, with repetition 6³ = 216.
How many four-digit numbers without repeated digits can be made from 0, 1, 2, 3, 4? The first digit cannot be 0: 4 ways; for the next places there are 4 remaining digits (0 included), then 3, then 2. Total 4 · 4 · 3 · 2 = 96. If a sports club has 4 football groups and 3 swimming groups, joining exactly one group can be done in 4 + 3 = 7 ways (addition rule).
Class activity
“Letters of a word”: take a word with all different letters (for example KITOB). List all orders of 3 of its letters by hand and check that there are 3! = 6. Then compute 5! for all five letters with the formula.
Practice
1
There are 5 roads from town A to B and 3 roads from B to C. In how many ways can one go from A to C?
15
2
In how many ways can the gold, silver and bronze medallists be chosen from 6 runners?
120
3
In how many ways can 6 differently coloured flags be hung in a row?
720
4
Why are the numbers of ways multiplied, not added, for consecutive choices?
Each result of the first choice forms a pair with every result of the second choice. So each of the m results of the first is repeated n times: n + n + … + n (m times) = mn.