Combinatorics DPP - RACE 4 - Selection of one or more objects

Selection Theory

Q1: Find the number of selection that can be made from 5 green balls, 4 blue balls and 3 red balls, if at least 1 green and 1 blue ball is to be included such that:

(i) All balls are different:
Total: 2^12
Only Red balls: 2^3 
Only Red,Green: 2^3.2^5 = 2^8
Only Red,Blue: 2^3.2^4 = 2^7 
Final: 2^12 + 2^3 - 2^8 - 2^7  = 3720

Note that all 3 sub-cases above include the case when no ball is selected.
But since we are subtracting Only red case from other two cases, that gets handled as well.
(ii) All balls of same color are identical:
5*4*4 = 80

Q2. Find the number of ways of selection of one or more letters from the letters of the word ‘INDEPENDENCE’ if

(i) There is no restriction.
Total letters:12
1I,3N,2D,4E,1P,1C
2*4*3*5*2*2 - 1 = 480 - 1 = 479 

(ii) The letters D & E are selected at least once.
2*4*2*4*2*2 = 256 

(iii) Only one letter is selected.
6C1 = 6

(iv) At least two letters are selected.
Total - 0 letters - 1 letter
= 480 - 1 - 6 = 473

Q3. In how many ways at least one vowel and at least one consonant can be selected from the letters of the word ‘BETWEEN’:
Total letters: 7
1B, 3E, 1T, 1W, 1N
At least one vowel: 3 ways(1E, 2E or 3E)
At least one consonant: 2^4 - 1 = 15
Total: 3*15 = 45

Q4. Let N = 200200, then:

(i) Find total number of divisor of N:
Factors: 11 * 13 * 7 *2^3 * 5^2
2*2*2*4*3 = 96 

(ii) Find total number of proper divisor of N
96 - 1 = 95

(iii) Find total number of odd divisor of N

2*2*2*1*3 = 24 

(iv) Find total number of even divisor of N

2*2*2*3*3 = 72 

(v) Find total number of divisor of N which are divisible by 10

2*2*2*3*2 = 48

(vi) Find total number of divisor of N which are divisible by 10 but not 14
2*2*1*3*2 = 24

(vii) Find total number of divisor of N which are neither divisible by 10 nor 14
Total: 2*2*2*4*3 = 96
Div by 10: (At least one 5 and at least one 2): 2*2*2*3*2 = 48
Div by 14: (At least one 7 and at least one 2) 2*2*1*3*3 = 36 
Div by both 10,14 (At least one 5, one 7 and at least one 2): 2*2*1*3*2 = 24
Answer: Total - divBy10 - divBy14 + divByBoth10And14
= 96 - 48 - 36 + 24 + 36

Another way:
When at least one 2 is there and 5 and 7 are missing: 2*2*3 = 12
When 2 is not there and 5 and 7 can be chosen freely: 3*2*2*2 = 24

(viii) Find total number of divisor of N which are perfect square.
First we have to identify all those factors which have power > 1:
2^4, 5^2
Now divide their powers by 2 and take floor:
[3/2] = 1 => There are 1 + 1 = 2 ways to select powers of 2, i.e. 2^0,2^2
Similarly for 5:
[2/2] = 1 => There are 1 + 1 = 2 ways to select powers of 5, i.e. 5^0,5^2
So there are 2*2 = 4 ways to select perfect squares.

(ix) Find the number of divisors of the form 2y+1, where y is a natural number.
If we put y = 1,2,3... We get factors like 3,5,7,9...
It means that all the odd factors with the exception of 1.
So we can not select any 2.
2*2*2*1*3 - 1 = 23.

(x) Find sum of all divisor of N
(7^0 + 7^1)(11^0 + 11^1)(13^0 + 13^1)(5^0 + 5^1 + 5^2)(2^0+2^1+2^2+2^3)
= 8 * 12 * 14 * 31 * 15 = 624960

(xi) Find the sum of all divisor divisible by 10
(7^0 + 7^1)(11^0 + 11^1)(13^0 + 13^1)(5^1 + 5^2)(2^1+2^2+2^3)
= 8 * 12 * 14 * 30 * 14 = 564480

(xii) Find the sum of all divisor divisible by 35
(7^1)(11^0 + 11^1)(13^0 + 13^1)(5^1 + 5^2)(2^0 + 2^1+2^2+2^3)
= 7 * 12 * 14 * 30 * 15 = 529200

(xiii) In how many ways N can be expressed as the product of two factors.
Each divisor effectively creates 2 factors.
Let's take N = 15
Divisors: 1,3,5,15
Factors: 1,15 | 3,5 | 5,3 | 15,1
And each factor pair is repeated twice.
So Total divisors/2 = 48.

(xiv) In how many ways N can be expressed as the product of two relatively prime factors.
Let's take N = 36
Factors: 2,2,3,3
Total divisors: 3*3 = 9
Divisors: 1,2,4,3,9,6,18,12
Relative prime factor pairs:
1,12
4,9
Total: 2


Let's take N = 180 = 2^2*3^2*5
Relative prime factor pairs:
1,180
4 ,45
9,20
5,36
Total: 4

So the logic is this. We are dividing the distinct factors into 2 groups. 
Same number can't be present in both the groups.
Let's say ab = N
So each distinct number can belong to either a or b.
So total: 2^5 choices.
But this will count each (a,b) twice.
So 2^5/2 = 16.

Another way to see this is:
We have to choose 0,1,2,3,4,5 factors:
5C0 + 5C1 + 5C2 + 5C3 + 5C4 + 5C5
But 5C0, 5C5 are doing the same thing.
Similarly others.
So divided the total by 2.
Total is (1+1)^5 = 2^5
Divide by 2: 2^4 = 16

Q5. Let m denote the number of ways in which 4 different books can be distributed among 10 persons, each receiving at most one and let n denote the number of ways of distribution if the books are all alike, Then m/n =
Solution:
m = 10C4.4!
n = 10C4
m/n = 4! = 24

Q6: A set contains 11 distinct elements, then the total number of subsets of the set having at least three elements is:
Solution:
Total subsets: 2^11
Subsets with 0 elements: 1
Subsets with 1 elements: 11
Subsets with 2 elements: 11C2
So final answer: 2048 - 67 = 1981

A wrong way to solve this:
Choose 3 elements first: 11C3
Remaining 8 elements may or may not be chosen: 2^8
So 2^8*11C3
What's wrong with this?
Let's say we initially chose {a,b,c} and then at a later step chose 'd' out of the remaining 8 to make {a,b,c,d}.
Now let's say we initially chose {b,c,d} and then at a later step chose 'a' out of the remaining 8 to make {a,b,c,d}.
So this set will be counted multiple times.

Q7: If fruits of same species are alike, then in how many ways at least 2 fruits can be selected from 5 Apples, 4 Mangoes, 3 Bananas and 3 different fruits?
Solution:
Total ways: 6*5*4*2*2*2 = 960
0 fruits: 1
1 fruit: 6C1 = 6
Answer: 960 - 7 = 953

Q8. Product of all even divisors of 1000 is:
Solution:
Factors: 2^3,5^3
Each even factor will have at least one 2.
So total number of ways to choose divisors: 3*4 = 12
2^1 will occur 4 times, once each with 5^0,5^1,5^2,5^3
Same for 2^2 and 2^3.
If we multiply all we will get: 2^(1*4).2^(2*4).2^(3*4) = 2^24
Similarly for 5: 5^(0*3).5^(1*3).5^(2*3).5^(3*3) = 5^18
Answer: 2^24.5^18

Q9. For each positive integer k, let Sk denote the increasing arithmetic sequence of integers whose first term is 1 and whose common difference is k. Then the number of values of k for which Sk contains the term 361 is:
Solution:
1 + (n-1)k = 361
(n-1)k = 360
Let (n-1) = m then
mk = 360
So we factorize 360 = 2^3.3^2.5
So number of distinct divisors of 360 are the distinct values of k.
So 4*3*2 = 24 is the answer.





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