Continued permutations

So we have just embarked into the world of permutations. A world where order does matter and we are counting the number of arrangements. We have learnt the new notation and got an idea on both diagramatic and direct mathematical calculations of them.

This set just looks at slightly more difficult questions and questions where we impose some restrictions on the arrangements.  It's still a relatively simple process to identify what you are looking for, especially if you start with a simple diagramatic representation first. 

I'll show you what I mean. 

Example 1

Suppose I have $5$5 colours, Red, Blue, Pink, Black and White to choose from to make a flag.  I am allowed to select $3$3 colours but I MUST have Red because it is the King's favourite colour. 

My diagramatic representation looks like this.  I start with $3$3 boxes (as I can choose three colours) 

I fill in the first box with RED, because I have to select it.

How many possible choices do I have for the second colour?  Well I have $4$4 (Blue, Pink, Black or White). 

And then how many possible choices do I have for the third colour?  I have $3$3. 

And altogether I have $4\times3=12$4×3=12 possible combinations of colours.  

Example 2

From the digits $3,4,7,9$3,4,7,9 and $0$0 how many $4$4 digit numbers greater than $5000$5000 can be made?

 

Starting with $4$4 boxes (because we are creating $4$4 digit numbers)

The first box represents how many possible choices I have for the first digit. The first digit has to be either the $7$7 or the $9$9 because I want a number greater than $5000$5000.  So I have $2$2 choices for this position.

$2$2

 

The next $3$3 positions can be filled with any of the remaining numbers.  ($4$4 options, then $3$3 options then $2$2)

$2$2	$4$4	$3$3	$2$2

 

Which means the total possible arrangements are $2\times4\times3\times2=48$2×4×3×2=48 possible numbers. 

 

Example 3

Suppose I have the digits $1,3,4,7,8,9$1,3,4,7,8,9.  How many $3$3 digit odd numbers can be made?

I'll start with $3$3 boxes, and write a $4$4 in the final position because there are only 4 possibilities for this to make the number an odd number.  

		$4$4

 

After fulfilling this requirement I have $5$5 options for the first position and $4$4 for the second.

$5$5	$4$4	$4$4

 

This means there are $80$80 possible arrangements.  Without the restriction of the number needing to be odd, there would have been $120$120 arrangements. $(P(6,3)=6\times5\times4)$(P(6,3)=6×5×4)

 

More Worked Examples

QUESTION 1

QUESTION 2