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i need help in AS-LEVEL STATISTICS permutation &combination

multixamza01

multixamza01

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awsumness said:
plz can some one explain permutation s....
Click to expand...

A permutation an arrangement of objects in a definite order.
Take the following question for example :
Find the total number of different number of permutations of the letters of the word SMART.

What the question is actually asking is that in how many ways can we arrange these letters .

Let's solve :
Notice that all the letters are different. So we got 5 different letters.
The answer is 5! (five factorial) . => 5!= 120.

There are questions in which restrictions are given. So let's take an example.

Find the number of permutations of the letters in the word HISTORY.
Find the number of these permutations in which
a) the letters O and R are together
b) the letters O and R are not together.

Let's solve :

The permutations of 7 different letters = 7! = 5o40

a)Now draw 6 blocks .
Now that O and R are to be put together in one block . Now we dont know whether the arrangement is OR or RO. So we take this permutation to be 2! . The remaining 5 blocks will be filled by the remaining letters. And as a whole the 6 blocks can be permuted in 6! ways

according to the counting principle = 2! x 6! =1440

b) 5040- 1440 =3600.
 
XPFMember

XPFMember

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Assalamoalaikum wr wb! :)

Check this post of mine..

XPFMember said:
For permutations and combinations watch this whole series of the tutorial, and you're good to go! ;) http://www.examsolutions.co.uk/maths-re ... rial-1.php


I see people finding permutations and combinations hard, but honestly i enjoy it a lot..

And one more thing, (you may understand this better may be after you've gone through the topic, but anyways)
to make out when you have to use permutation (nPr of the calculator) and when to use combinations (nCr) , checck if the different arrangements of the selections will make any difference or not...if it does then you've to use nPr, and if it doesn't then u must use nCr.

Eg. lets say I want to choose two numbers from a total of three..how many possible selections are there?

now see i may choose 1 2 ; 1 3 ; 2 3
now the no. i've chosen can be written as 2 1 ; 3 1 ; 3 2

and obviously, this is giving me different numbers, so here arrangement does matter..so go for nPr => 3P2 = 6

but say if i got to chose 2 people from a total of 3 (A, B and C)

so possibilities are A B ; A C ; B/ C
and if i make this as B A ; C A ; C B
it doesn't make any difference as in the end there are the same people whether A first or B first....so that means arrangement does not matter...we'll use nCr => 3C2 = 3

P.S. These were pretty simple examples, I used them to explain the concept...you'll find this part of my explanation helpful when later on you tackle some more harder situations...this part really makes this thing easy...it did for me ;)
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