1 - Lucky or Unlucky?
The summer vacation had been a boring time for Kabani. She was not allowed to go out and play in the sun, she was in her room all day painting. Even painting was getting a bit boring now... Few of her mother's friends came home, they had been talking long.
Kabani overheard one of her aunts saying "1 is so unlucky for me!". She wondered how it really mattered. Kabani started counting how many other digits were there - 0, 2, 3, 4, 5, 6, 7, 8 and 9 - i.e., nine of them! Then she started counting the number of two-digit numbers which didn't have 1 - 00, 02, 03, 04... - she finally had written down all of them. There were 81 one of them. The most interesting observation she made was that each digit could be followed by any of the nine digits, which meant she could have 9*9 (=81) possible two-digit numbers without 1.
Taking the calculation forward, she predicted that there will 9*9*9 = 729 three-digit numbers without 1 and there would be 9*9*9*9 = 6,561 of those four-digit numbers! Kabani wondered how these numbers would change if her aunt felt that both 1 and 2 were unlucky. The answers were simple, there would be 8 one-digit numbers, 8*8 = 64 two-digit numbers, 8*8*8 = 512 three-digit numbers and 8*8*8*8 = 4,096 four-digit numbers which don't contain 1 or 2!
Permutations with Repetitions
The above problem falls into the following class of problems. There are n different objects. We have to select objects from the above n to fill r vacancies in a straight line. When each of the r vacancies has been filled, we call it an arrangement (also called as r-arrangements). Each r-arrangement can contain more than one object of the same type (this is referred to as repetition) and two r-arrangements would be considered distinct if they vary at least at one of the r positions. The task is to find the total number of such distinct arrangements. These arrangements are called r-permutations of n distinct objects with repetitions. For filling the first vacancy we have n options, for filling the second vacancy we have n options, for the third it is n options and so on. So for filling each of the r vacancies we will have n options and hence the total number of ways we can make the arrangement is nr . It would also be equal to the number of distinct r-arrangement possible.
In the earlier problem, we had to find the number of possible two-digit numbers from 9 distinct digits. The answer would be 92 ! All the calculations of Kabani can now be done easily without wasting much time and energy. Permutations with repetitions have many valuable applications in the world around us, we will explore a few of them in the next article.
Kabani overheard one of her aunts saying "1 is so unlucky for me!". She wondered how it really mattered. Kabani started counting how many other digits were there - 0, 2, 3, 4, 5, 6, 7, 8 and 9 - i.e., nine of them! Then she started counting the number of two-digit numbers which didn't have 1 - 00, 02, 03, 04... - she finally had written down all of them. There were 81 one of them. The most interesting observation she made was that each digit could be followed by any of the nine digits, which meant she could have 9*9 (=81) possible two-digit numbers without 1.
Taking the calculation forward, she predicted that there will 9*9*9 = 729 three-digit numbers without 1 and there would be 9*9*9*9 = 6,561 of those four-digit numbers! Kabani wondered how these numbers would change if her aunt felt that both 1 and 2 were unlucky. The answers were simple, there would be 8 one-digit numbers, 8*8 = 64 two-digit numbers, 8*8*8 = 512 three-digit numbers and 8*8*8*8 = 4,096 four-digit numbers which don't contain 1 or 2!
Permutations with Repetitions
The above problem falls into the following class of problems. There are n different objects. We have to select objects from the above n to fill r vacancies in a straight line. When each of the r vacancies has been filled, we call it an arrangement (also called as r-arrangements). Each r-arrangement can contain more than one object of the same type (this is referred to as repetition) and two r-arrangements would be considered distinct if they vary at least at one of the r positions. The task is to find the total number of such distinct arrangements. These arrangements are called r-permutations of n distinct objects with repetitions. For filling the first vacancy we have n options, for filling the second vacancy we have n options, for the third it is n options and so on. So for filling each of the r vacancies we will have n options and hence the total number of ways we can make the arrangement is nr . It would also be equal to the number of distinct r-arrangement possible.
In the earlier problem, we had to find the number of possible two-digit numbers from 9 distinct digits. The answer would be 92 ! All the calculations of Kabani can now be done easily without wasting much time and energy. Permutations with repetitions have many valuable applications in the world around us, we will explore a few of them in the next article.
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