practice problems pending

Q1. Find the number of ways to choose an ordered pair ((a,b)) of numbers from the set (1...10) such that (|a-b| <= 5).
S1.
Answer: 80
Method 1: Direct counting
1: 1,2,3,4,5,6 => count:6
Similarly the count is 6 for 2,3,4,5.
So total: 30*2 = 60 pairs.
Then for 6,7,8,9,10 there are 5,4,3,2,1 options.
So 15*2 = 30 pairs.

Total: 90
But we have counted same number pairs twice.
So 90-10=80 = answer

Method 2: faster and better: complementary counting
Total: 10x10 = 100
Subtract invalid:
1: 7,8,9,10
2: 8,9,10
3: 9,10
4: 10

Total: 10*2 = 20
100-20 = 80 = Answer

Q2. Suppose that in a poll made of 150 people, the following information was obtained: 70 of them read The Hindu, 80 read The Indian Express and 50 read Deccan Herald. 30 read both The Hindu and The Indian Express; 20 read both The Hindu and Deccan Herald and 25 read both The Indian Express and Deccan Herald. Find at most how many of them read all the three.
S2.
Incorrect solution first:
Initially I applied PIE without thinking well.
I thought 150 = |H U I U D|
=> 150 = 70 + 80 + 50 - 30 - 20 - 25 + |H Intersect I Intersect D|
=> |H Intersect I Intersect D| = 25

But this is mistaken.
Why?
The question didn't mention the number of those who don't read any newspaper at all.
Also the question did give a hint by saying "at most how many".

Also, if we look at the pairwise intersection counts: 30,20,25, it's clear that number of people reading all 3 can't be more than 20.

Here is the correct approach:
So just fixing our initial approach:
|H U I U D| = 70 + 80 + 50 - 30 - 20 - 25 + |H Intersect I Intersect D|
=> 125 + |H Intersect I Intersect D| = |H U I U D|
But |H U I U D| <= 150 since 150 is total number of people polled.
=> 125 + |H Intersect I Intersect D| <= 150
=> |H Intersect I Intersect D| <= 25

But as mentioned earlier, it has to be less than the pairwise intersections
=> |H Intersect I Intersect D| <= 20 = answer.

Q3. Lewis Carroll, the famous author of Alice in Wonderland, Through the Looking Glass, The Hunting of the Shark and other wonderful works, was a mathematician whose real name was Charles Lutwidge Dodgson (1832–1898). Here is a problem from his book A Tangled Tale.

Let (S) be the set of pensioners, (E) the set of those who lost an eye, (H) those who lost an ear, (A) those who lost an arm and (L) those who lost a leg.

Given that (n(E)=70%), (n(H)=75%), (n(A)=80%) and (n(L)=85%). Find what percentage at least must have lost all the four.

S3.
Use complements.
E' = number of people who didn't lose an eye = 30 (assume that in total we have 100 people)
H' = 25
A' = 20
L' = 15

Number of people who didn't lose all 4 belong to (E' U H' U A' U L').
To minimize the people who lost all 4, we need to maximize the number above.
For that all of the complement sets should be mutually exclusive.
=> Maximum number of people who didn't lose all 4 = 30 + 25 + 20 + 15 = 90
=> Minimum number of people who lost all 4 = 100 - 90 = 10% = Answer

Q4. In the above problem, if those who lost all the four are more than 10 and less than 70, construct an example.
S4.
Let Minimum number of people who lost all 4 = 15
=> Max people who didn't lose all 4 = 85
Let E' intersection H' be 5
So Person number 1 to 30 in E'.
From 26 to 50 in H'
51 to 70 in A'.
71 to 85 in L'.

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