k-permutations - Exercise set 1

This exercise set contains some solved exercises on k-permutations. The theory needed to solve these exercises is introduced in the lecture entitled k-permutations.

Exercise 1.1

There is a basket of fruit containing an apple, a banana and an orange and there are five girls who want to eat one fruit. How many ways are there to give three of the five girls one fruit each and leave two of them without a fruit to eat?

Solution

Giving the 3 fruits to 3 of the 5 girls is a sequential problem. We first give the apple to one of the girls. There are 5 possible ways to do this. Then we give the banana to one of the remaining girls. There are 4 possible ways to do this, because one girl has already been given a fruit. Finally, we give the orange to one of the remaining girls. There are 3 possible ways to do this, because 2 girls have already been given a fruit. Summing up, the number of ways to assign the three fruits is equal to the number of 3-permutations of 5 objects (without repetition). If we denote it by $P_{5,3}$ , then: [eq1]

Exercise 1.2

An hexadecimal number is a number whose digits can take sixteen different values: either one of the ten numbers from 0 to 9, or one of the six letters from A to F. How many different 8-digit hexadecimal numbers are there, if an hexadecimal number is allowed to begin with any number of zeros?

Solution

Choosing the 8 digits of the hexadecimal number is a sequential problem. There are 16 possible ways to choose the first digit and 16 possible ways to choose the second digit. So, there are 16x16 possible ways to choose the first two digits. There are 16 possible ways two choose the third digit and 16x16 possible ways to choose the first two. Thus, there are 16x16x16 possible ways to choose the first three digits. An so on, until we have chosen all digits. Therefore, the number of ways to choose the 8 digits is equal to the number of 8-permutations with repetition of 16 objects:

Exercise 1.3

An urn contains ten balls, each representing one of the ten numbers from 0 to 9. Three balls are drawn at random from the urn and the corresponding numbers are written down to form a 3-digit number, writing down the digits from left to right in the order in which they have been extracted. When a ball is drawn from the urn it is set aside, so that it cannot be extracted again. If one were to write down all the 3-digit numbers that could possibly be formed, how many would they be?

Solution

The 3 balls are drawn sequentially. At the first draw there are 10 balls, hence 10 possible values for the first digit of our 3-digit number. At the second draw there are 9 balls left, hence 9 possible values for the second digit of our 3-digit number. At the third and last draw there are 8 balls left, hence 8 possible values for the third digit of our 3-digit number. Summing up, the number of possible 3-digit numbers is equal to the number of 3-permutations of 10 objects (without repetition). If we denote it by $P_{10,3}$ , then: [eq3]

by Marco Taboga

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