For these problems, providing an answer is not enough: you must specify what are
ID: 675291 • Letter: F
Question
For these problems, providing an answer is not enough: you must specify what are the sets that you are counting and why you can use the sum or the product principle, and other arguments that may be necessary. How many subsets does the set {1, 2,...,n} have? Prove your answer using the Product Rule. Using the digits 1, 3, 5 and 7, how many 4 digit numbers can be formed if The first digit must be 1 and repetition of the digits is not allowed? The number must be divisible by 5 and repetition is allowed? The number must be divisible by 5 or the first digit must be 1? A hiring committee must be formed of four people from the Department of Computer Science. There are 9 Faculty members that are eligible for the committee. How many possible committees are there? In how many ways can we arrange 10 people into 5 groups of size 2?Explanation / Answer
1. Set {1,2,......N}
number of subsets : Please note for every set with N elements,,, No of subsets are 2^n
for example set {1,2} can have 2^2 subsets = 4 subsets
{1,2}= {1}, {2}, {1,2}, { }(null set)
Similarly
{1,2,3}= {1}{2}{3}{1,2}{1,3}{2,3}{1,2,3}{ } = 2^3= 8 Sets
Similarly set {1,2,,,,,,N} = 2^N sets.
Problem 14.2
a. numbers are 1,3,5,7
so We have four places to fill wth four numbers
1 _ _ _ and first digit is fixed with 1. So put 1 at first place and repeition is not allowed.
no of choices: 3*2*1 // after fixing 1 at first place , second place has only 3 choices(3,5,7) after this third place left only with 2 choice because repetition is not allowed and after that last place has left with only 1 choice.
So total no of words are : 3*2*1= 6 numbers.
B. No must be divisible by 5 and repition is allowed.
number to be divisible by 5 must contain 5 at its unit place. So fix 5 at unit place.
_ _ _ 5 // No of choices left for other digits.
no of choices : 3*3*3 (here repition is allowd thats why we have 3 choices for every place)= 27 Words.
C. No must be divisible by 5 or first digit must be 1
In this case add result of above two cases. So 6+ 27 = 33 .
Problem 14.3 : 4 people out of 9 eligible people
So answer is simple we have to use Combinations not permutations. Combinations would be 9C4= 9!/(4! *5!)
Please note Here permutation will not be used because we not require arrangement here because even after arrangement people would remain same here.
Problem 14.4
10!/(2!*2!*2!*2!*2!*5!)
Multiplying these factors:
(10C2)*(8C2)*(6C2)*(4C2)*(2C2)
Now we need to divide that total by 5!, since the labeling/ordering of the groups (group ONE, group TWO, group THREE group FOUR, Group FIVE) doesn't matter: since there are 5! ways of labeling the groups, we need divide the total by 5!
So final Answer would be
10!/(2!*2!*2!*2!*2!*5!)