Unizor  - Combinatorics - Simple Permutations Problems 1
Channel: Zor Shekhtman on UNIZOR Education
Date: 20/05/2014

Don't try to recall any formula. Instead, use the logic and count available choices.

Problem 1
Using the logic of permutations, determine how many 3-digit numbers are there between 500 and 799 (inclusively).
Answer
There are 3 choice for the first digit (5, 6 and 7), 10 choices for the second (from 0 to 9) and 10 choices for the third (from 0 to 9) making the overall number of choices equal to
3·10·10 = 300
This corresponds to a direct approach based on subtraction since this number should be equal to
800 − 500 = 300

Problem 2
Using the logic of permutations, determine how many even 3-digit numbers are there between 500 and 799 (inclusively).
Answer
There are 3 choice for the first digit (5, 6 and 7), 10 choices for the second (from 0 to 9) and 5 choices for the third (0, 2, 4, 6, 8) making the overall number of choices equal to
3·10·5 = 150
This corresponds to a direct approach based on subtraction since this number should be equal to
(800 − 500) / 2 = 150

Problem 3
Three cards are selected from a standard deck of 52 cards by shuffling a deck and picking the top three cards.
How many different permutations of a deck exist when the top three cards are 3, 7 and an ace (in that order)?
Answer
There are 4 choices for the first card (3 of clubs, of diamonds, of hearts and of spades), 4 choices for the second card (7 of clubs, of diamonds, of hearts and of spades) and 4 choices for the third card (ace of clubs, of diamonds, of hearts and of spades).
For the 4th card we have a choice of 49 cards (any one that is left after picking the first three cards).
For the 5th card we have a choice of 48 cards (any one that is left after picking the first four cards).
The number of choice goes down to 1 one-by-one with every next pick.
The result is
4·4·4·49!

Problem 4
There are 10 participants in the competition. Top three get the medals (gold, silver and bronze).
How many different results of the medal distribution among participants can you count?
Answer
This is a typical problem on partial permutations. We have to pick 3 participants out 10 in a specific order. For the gold medal we have 10 choices, for the silver - remaining 9 and for the bronze - remaining 8. The total number of distributions of three medals among 10 participants is
10·9·8 = 720

Problem 5
A gardener has 20 rows where he has to plant flowers. He does not want any two rows to be identical in a sense of the order of flowers.
What is the minimum number of different kinds of flowers does he need to satisfy this requirements?
Answer
The number of different kinds of flowers N should be sufficient in a sense that the total number of permutations of N objects (that is, N!) should be greater or equal to 20. The number N=3 is insufficient since 3! = 6 is less than 20. But number N=4 should suffice since 4! = 24 is greater than 20.
So, the answer is: a gardener needs at least 4 different kinds of flowers in order to make different 20 rows.

Problem 6
After graduation from a medical school a student sends the documents to the agency that helps to find him a hospital to go through the residency program. There are 20 different hospitals to choose from, but the student has to specify only two of them: choice #1 and choice #2 hospitals.
How many different decisions about the residency choices are possible to make?
Answer
This is a typical partial permutation problem (as the problem 4 above). For the hospital #1 there are 20 different choices. With each of them there are 19 choices left for the hospital #2. That gives the total number of possible decisions
20·19 = 380