How to solve a Permutation problem.

"The essence of mathematics is not to make simple things complicated, but to make complicated things simple."

S. Gudder

One of the most "amusing" topics in add maths would be "permutations and combinations". The problems are normally amongst the most easily confused and hard to solve. Some questions are worded in the way that the more you read the questions the worse it gets, in term of understanding the problems. And some students just wondered why do we need to bother thinking how to arrange the words FIELD in a row with various arrangements or combinations or having a hard time figuring out how many odd numbers can be formed using a set of given digits!! :D

In order to understand permutations and combinations, you must appreciate the use in real life. Well, supposed your teacher wants to split your class of 40 into 5 groups to work on a project, how many ways can he do it?

You may wonder sometimes how many kinds of combinations you can make out of having three tudung ( head cover), five blouses and three skirts?

Or if you are arranging a number of books on two level shelf with various specifications ( e.g. all the history books next to each other, all form five books are in the bottom shelf, etc)


Okay enough said about the use and applications in real life. Let's have a look at a question that came from a student who was doing an exercise using some exercise book.

Question

How many different odd numbers can be formed using the digits 3, 4,5, 6 and 7? How many of these odd numbers are more than 50,000?

A reader may easily fall into thinking or interpreting that there are two digits odd number or three digits or four digits to be counted in and of course some limit only to count in those with ALL five digits.

If you are faced with any permutation and combination questions, read the sentence carefully. In this case it sounded more like it is interpreted as ALL given digits must be used.

Part 1 - For the last digit, to make it an odd numbers the end of each numbers must be either 3,5 or 7.

Therefore 3P1 ways of using either one of these.

The other four digits will be taken from the other left overs ( 4 balanced digits) : 4P4

Hence, the arrangement of odd number using the digits: 4P4 x 3P1 = 72 ways.


Part 2 : How many of these odd numbers are more than 50,000.

There are two conditions to be met in this part. First, the first digit will either be number 5,6 or 7 AND the last digit will either be 3,5 or 7.


For the numbers to be more than 50,000 the first digit must be either 5,6 or 7.

If the first digit 5 or 7 ( more than 5 and odd number) then your last digit will be affected. Note that your last digit must be either 3,5 or 7.

And you can only arrange the first digit 2P1 ways.

But for the last digit it must be picked from the odd numbers to make them odd ( either 3, 5 or 7) , yet one (either 5 or 7) would have been use as the first digit. Whichever is chosen, you only have 2 more to choose from, which makes it 2P1 for the last digit.

Now, you are left with three more digits in between those first and last digits. You can arrange them in 3P3 ways.

Multiply all these arrangements

Hence, 2P1 x 3P3 x 2P1 = 24

If first digit is 6 ( more than 50,000 but an even number) your last digit will have three choices ( 3,5,7) as none of the odd numbers would be used in the first digit position. 

The first digit can be arranged in 1P1

Your last digit can be chosen from either 3,5,7 ( odd numbers) = 3P1

Your three other digits will be arranged in 3P3 ways.

Hence you can make 1P1 x 3P3 x 3P1 = 18 ways

Grand total: 24 ways + 18 ways = 42 ways.