Unizor - Combinatorics - Simple Combinations Problems 2

Problem 1
There are 6 vacancies in the company. 3 of them require skill A, 2 vacancies requires skill B and 1 vacancy does not require any specific skill, so anybody can fill this vacancy.
There are 8 applicants. 4 of them have only skill A and the other 4 have only skill B.
How many different ways to fill all 6 vacancies are there?
Solution
C(4,3)·C(4,2)·C(3,1) = 72

Problem 2
How many different ways are there to represent the number 6 as a sum of three non-negative integer numbers? Consider a sum of the same three numbers but in a different order as a different representation.
Solution
C(8,2) = 8!/(2!·6!) = 8·7/2 = 28

Problem 3
A new Olympic team of 5 members has to be chosen out of 10 top participants in the national championship. Among them are the gold and the silver medal winners of the prior Olympic games who automatically must be in the team. How many different teams can be chosen?
Solution
C(8,3) = 8!/(3!·5!) = 56

Problem 4
There are 20 participants in the tennis tournament. Each participant should play with each other only one match.
How many matches will be played?
Solution
C(20,2) = 20! / (2!·18!) = (20·19)/(1·2) = 190

Problem 5
Consider all 6-digit numbers. Some of them have at least one digit 9 among these 6 digits, some not. Which 6-digit numbers are more numerous, those that have digit 9 among its digits or those that don't?
Solution
Number of numbers with no 9's:
8·9·9·9·9·9 = 472,392
Number of all numbers in the interval:
900,000 numbers in between.
As you see, we have slightly more than half of 6-digit numbers without digit 9 than with it.
Problem 6
Let's consider all numbers from 0 up to the maximum number that can be written with N digits (i.e. 99...9 where the number of digits equals to N). Calculate the ratio of numbers in this interval that do not contain the digit 9 relative to the total number of numbers. Will it ever be the case that the numbers without digit 9 become a minority, so the ratio becomes smaller than one half?
Solution
9^N / 10^N = 0.9^ N
Straight verification shows that
0.9^6 = 0.531441 and
0.9^7 = 0.478297.
Therefore, among numbers not exceeding 6 digits the numbers without digit 9 are the majority, but, if numbers up to 7 digits are considered, they are already a minority and there are more numbers with digit 9 among them than without.

[ZorShekhtman]

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[ZorShekhtman]

Problem 1
There are 6 vacancies in the company. 3 of them require skill A, 2 vacancies requires skill B and 1 vacancy does not require any specific skill, so anybody can fill this vacancy.
There are 8 applicants. 4 of them have only skill A and the other 4 have only skill B.
How many different ways to fill all 6 vacancies are there?
Solution
C(4, 3)·C(4, 2)·C(3, 1) = 72

Problem 2
How many different ways are there to represent the number 6 as a sum of three non-negative integer numbers? Consider a sum of the same three numbers but in a different order as a different representation.
Solution
C(8, 2) = 8!/(2!·6!) = 8·7/2 = 28

Problem 3
A new Olympic team of 5 members has to be chosen out of 10 top participants in the national championship. Among them are the gold and the silver medal winners of the prior Olympic games who automatically must be in the team. How many different teams can be chosen?
Solution
C(8, 3) = 8!/(3!·5!) = 56

Problem 4
There are 20 participants in the tennis tournament. Each participant should play with each other only one match.
How many matches will be played?
Solution
C(20, 2) = 20! / (2!·18!) = (20·19)/(1·2) = 190

Problem 5
Consider all 6-digit numbers. Some of them have at least one digit 9 among these 6 digits, some not. Which 6-digit numbers are more numerous, those that have digit 9 among its digits or those that don't?
Solution
Number of numbers with no 9's:
8·9·9·9·9·9 = 472, 392
Number of all numbers in the interval:
900, 000 numbers in between.
As you see, we have slightly more than half of 6-digit numbers without digit 9 than with it.
Problem 6
Let's consider all numbers from 0 up to the maximum number that can be written with N digits (i.e. 99...9 where the number of digits equals to N). Calculate the ratio of numbers in this interval that do not contain the digit 9 relative to the total number of numbers. Will it ever be the case that the numbers without digit 9 become a minority, so the ratio becomes smaller than one half?
Solution
9^N / 10^N = 0.9^ N
Straight verification shows that
0.9^6 = 0.531441 and
0.9^7 = 0.478297.
Therefore, among numbers not exceeding 6 digits the numbers without digit 9 are the majority, but, if numbers up to 7 digits are considered, they are already a minority and there are more numbers with digit 9 among them than without.