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How many arrangements are there for the $8$ letters of the word VISITING?
Solution:
So there are $\binom{8}{8}$, meaning 8 choose 8. But this gives a $8$ letter set not $8$ letter words, therefore
$\binom{8}{8}\times 8!$
1) Is this logic correct?
2) How do I evaluate $\binom{8}{8}\times 8!$
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This section covers permutations and combinations.
Arranging Objects
The number of ways of arranging n unlike objects in a line is n! (pronounced ‘n factorial’). n! = n × (n – 1) × (n – 2) ×…× 3 × 2 × 1
Example
How many different ways can the letters P, Q, R, S be arranged?
The answer is 4! = 24.
This is because there are four spaces to be filled: _, _, _, _
The first space can be filled by any one of the four letters. The second space can be filled by any of the remaining 3 letters. The third space can be filled by any of the 2 remaining letters and the final space must be filled by the one remaining letter. The total number of possible arrangements is therefore 4 × 3 × 2 × 1 = 4!
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The number of ways of arranging n objects, of which p of one type are alike, q of a second type are alike, r of a third type are alike, etc is:
n! .
p! q! r! …
Example
In how many ways can the letters in the word: STATISTICS be arranged?
There are 3 S’s, 2 I’s and 3 T’s in this word, therefore, the number of ways of arranging the letters are:
10!=50 400
3! 2! 3!
Rings and Roundabouts
- The number of ways of arranging n unlike objects in a ring when clockwise and anticlockwise arrangements are different is (n – 1)!
When clockwise and anti-clockwise arrangements are the same, the number of ways is ½ (n – 1)!
Example
Ten people go to a party. How many different ways can they be seated?
Anti-clockwise and clockwise arrangements are the same. Therefore, the total number of ways is ½ (10-1)! = 181 440
Combinations
The number of ways of selecting r objects from n unlike objects is:
Example
There are 10 balls in a bag numbered from 1 to 10. Three balls are selected at random. How many different ways are there of selecting the three balls?
10C3 =10!=10 × 9 × 8= 120
3! (10 – 3)!3 × 2 × 1
Permutations
A permutation is an ordered arrangement.
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The number of ordered arrangements of r objects taken from n unlike objects is:
nPr = n! .
(n – r)!
Example
In the Match of the Day’s goal of the month competition, you had to pick the top 3 goals out of 10. Since the order is important, it is the permutation formula which we use.
10P3 =10!
7!
= 720
There are therefore 720 different ways of picking the top three goals.
Probability
The above facts can be used to help solve problems in probability.
Example
In the National Lottery, 6 numbers are chosen from 49. You win if the 6 balls you pick match the six balls selected by the machine. What is the probability of winning the National Lottery?
The number of ways of choosing 6 numbers from 49 is 49C6 = 13 983 816 .
Therefore the probability of winning the lottery is 1/13983816 = 0.000 000 071 5 (3sf), which is about a 1 in 14 million chance.
Lia W. I don't understand how to divide out the 2 letters that have already been chosen.
2 Answers By Expert Tutors
the first 2 are already chosen how many permutations of 2 letters are there from the 8 letters ? for the first letter you are choosing from 8 letters for the second letter you are choosing from 7 letters 8*7=56 2-letter permutations from the 8 letters you said order matters; assuming order matters with the first 2 letters... AB, AC, AD, AE, AF, AG, AH BA, BC, BD, BE, BF, BG, BH and so on for a total of 56 2-letter permutations that leaves 6 letters from which to choose 3 letters and these are permutations 6*5*4=120 3-letter permutations
each of these 120 permutation can be combined with each of the 56 2-letter permutations
you could have gotten the answer as follows...
if the first 2 letters are permutations then you have 5-letter permutations from the 8 letters:
if the first 2 letters don't have to be permutations then there are only 56/2=28 combinations for the first 2 letters combined with the 120 permutations for a total of 28*120=3360 arrangements
Karyn T. answered • 08/09/16
For the love of Math! Experienced Math tutor for grades 6-12
When order matters you are doing a Permutation problem. Combination problems have fewer outcomes for the same set up because order does not matter.
Think about the eight letters A B C D E F G H, if A and B (or any two letters) are already chosen then you want to know how many ways 6 letters can fill 3 spaces when order matters, in other words the letters F G H are considered 1 outcome with the first two letters and G F H is considered another individual outcome with the first two letters. If order didn't matter then F G H and G F H with the first two letters would be considered only one outcome.
6P3 = 6!/[(6-3)!] = (6x5x4x3x2x1)/(3x2x1) = (6x5x4) = 120 ways
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