☰ Contents · Mathematics

Combinatorics and Newton’s binomial

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
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Combinatorics problems

Textbook, Part 2: pp. 27–32
GoalLearn to solve counting problems with the addition and multiplication rules, arrangements and permutations.
New words
addition rule · qo‘shish qoidasimultiplication rule · ko‘paytirish qoidasiarrangement · o‘rinlashtirishfactorial · faktorial
Explanation

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?
2
In how many ways can the gold, silver and bronze medallists be chosen from 6 runners?
3
In how many ways can 6 differently coloured flags be hung in a row?
4
Why are the numbers of ways multiplied, not added, for consecutive choices?