practice problems pending

Q1. A DNA chain is composed of basic building blocks in the form of four chemicals, known by the symbols A,C,T,G. Consider three-letter chains consisting of these symbols (with or without repetitions). How many such chains are there?
S1.
4*4*4 = 64

Q2. Consider the set {a,b,c,d,e}. How many three-letter words can be made out of them, with or without meaning? How many of these will have at least one vowel in them? Answer these questions both for the case when repetitions of letters are allowed and for the case when repetitions of letters are not allowed.
S2.
2.1
With repetition: 5*5*5 = 125
Without: 5*4*3 = 60

2.2
With repetition:
Total words: 5^3 = 125
Without any vowels = 3^3 = 27
Answer = 125 - 27 = 98

Without:
Total: 5*4*3 = 60
Without vowels = 3*2*1
Answer = 60 - 6 = 54


Q3. A permutation (a1,a2,a3,a4,a5) of (1,2,3,4,5) is called heavy-tailed if a1 + a2 < a4 + a5. What is the number of heavy-tailed permutations?
S3.
Total such: 5! = 120
Let's fix 1,2 to a1,a2 => 2!*3! = 12
1,3 => 12
1,4 => a3,a4 from 2,3,5 = 2,5 3,5 => 2!*4 = 8
1,5 => 2,3,4 => 3,4 => 2*2 = 4
1 is done.

2,3 => 1,5 4,5 => 8
2,4 => 1,3,5 => 3,5 => 4
2,5 => 0
2 is done

3,4 3,5 => 0
That's it.
Total = 48

Method 2:
By symmetry
Permutations where a1 + a2 > a4 + a5 exactly equal those where a1 + a2 < a4 + a5.
So if we count where a1 + a2 = a4 + a5 and then subtract from total and divide by 2 we will get answer.
Equality happens at:
1 + 4 = 2 + 3 => 2!.2!.2! = 8
2 + 4 = 1 + 5
3 + 4 = 2 + 5

So 120 - 24 = 96
96/2 = 48

Q4. How many ways are there to split the integers from 1 to 14 into 7 pairs such that in each pair, the greater number is at least 2 times the lesser number?
S4.
For e.g.
7,14 6,12 ... 1,2 => This is the simplest split.

Also, note that the lesser number will always be from {1,2...7} and greater will be from {8,9...14}.
If not, then for e.g. if you pick 3,6 then {1,2,4,5,7} will pair with {8,9...14} and at least one pair will be incompatible for e.g. 11,12.

Now.
7,14 has to be fixed. No option. 12 numbers remain.
6 can come with 12,13.

Case 1:
6,12 => 10 numbers remain.
5 has 3 options 10,11,13 => 8 numbers remain since 7,14,6,12,5 and one from (10,11,13) are gone.
4 has 4 options 8,9 and 2 from (10,11,13) => 6 numbers remain.
3 has to choose from 3 remaining from (8,9,10,11,13) => 4 numbers remain.
2 has to choose from 2 remaining from (8,9,10,11,13) => 2 remain (1 of which is 1)
1 will pair with whatever is left.
 = 3*4*3*2 = 72

Case 2:
6,13 => 10 numbers remain.
5 has 3 options 10,11,12 => 8 numbers remain since 7,14,6,13,5 and one from (10,11,12) are gone.
4 has 4 options 8,9 and 2 from (10,11,12) => 6 numbers remain.
3 has to choose from 3 remaining from (8,9,10,11,12) => 4 numbers remain.
2 has to choose from 2 remaining from (8,9,10,11,12) => 2 remain (1 of which is 1)
1 will pair with whatever is left.
= 72

Total: 144

Q5. Let S be the set of permutations of the sequence 1,2,3,4,5 for which the first term is not 1. A permutation is chosen randomly from S. The probability that the second term is 2 is a/b in lowest terms. What is a+b?
S5.
First term is not 1 => |S| = 4*4*3*2*1 = 96
First term is not 1 and second term is 2 => 3*1*3*2*1 = 18
a/b = 18/96 = 3/16
a+b = 19

Q6. Suppose 28 objects are placed along a circle at equal distances. In how many ways can 3 objects be chosen from among them so that no two of the three chosen objects are adjacent nor diametrically opposite?
S6.
Prereq: Circular adjacency non consecutive combinatorics



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