Permutation and Combinations Help S1 Watch

Fruitbasket786
Badges: 11
Rep:
?
#1
Report Thread starter 4 years ago
#1
So this a question I came across when doing some past paper questions and I know that permutations and combinations is one of the most difficult topics I think in the whole of the stats 1 paper.

The thing which made me confused was the fact that once you choose one possible combination what do you do with the other white and grey cards. I don't know how to find the combination involving all cards and not specific ones. Since I don't know how to do the first part I can't to the second part of it because of this.
Attached files
0
reply
TeeEm
Badges: 19
Rep:
?
#2
Report 4 years ago
#2
(Original post by Fruitbasket786)
So this a question I came across when doing some past paper questions and I know that permutations and combinations is one of the most difficult topics I think in the whole of the stats 1 paper.

The thing which made me confused was the fact that once you choose one possible combination what do you do with the other white and grey cards. I don't know how to find the combination involving all cards and not specific ones. Since I don't know how to do the first part I can't to the second part of it because of this.
order matters in part i
0
reply
poorform
Badges: 10
Rep:
?
#3
Report 4 years ago
#3
There are n! ways to rearrange n elements. Because there is n choices for the 1st number n-1 for the second....2 choices for the n-1th position and 1 choice for the last that is

n*n-1*n-2*...*3*2*1 possible rearrangements.

So there are 7! rearrangements of the 7 digits but you need to consider that having the number begin with 33 for example is the same as having it beginning with 33 again but the two 3's switched so you need to divide out the extra rearrangements that are the same.

How many ways are there to rearrange the 3 3's and the same with the 2 5's?

Can you see how this would help?
1
reply
Fruitbasket786
Badges: 11
Rep:
?
#4
Report Thread starter 4 years ago
#4
(Original post by poorform)
There are n! ways to rearrange n elements. Because there is n choices for the 1st number n-1 for the second....2 choices for the n-1th position and 1 choice for the last that is

n*n-1*n-2*...*3*2*1 possible rearrangements.

So there are 7! rearrangements of the 7 digits but you need to consider that having the number begin with 33 for example is the same as having it beginning with 33 again but the two 3's switched so you need to divide out the extra rearrangements that are the same.

How many ways are there to rearrange the 3 3's and the same with the 2 5's?

Can you see how this would help?
I know the concept of part one, I mostly don't get part 2

Posted from TSR Mobile
0
reply
X

Quick Reply

Attached files
Write a reply...
Reply
new posts
Latest
My Feed

See more of what you like on
The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

Personalise

University open days

  • Cardiff Metropolitan University
    Undergraduate Open Day - Llandaff Campus Undergraduate
    Sat, 27 Apr '19
  • University of East Anglia
    Could you inspire the next generation? Find out more about becoming a Primary teacher with UEA… Postgraduate
    Sat, 27 Apr '19
  • Anglia Ruskin University
    Health, Education, Medicine and Social Care; Arts, Humanities and Social Sciences; Business and Law; Science and Engineering Undergraduate
    Sat, 27 Apr '19

Have you registered to vote?

Yes! (502)
37.86%
No - but I will (101)
7.62%
No - I don't want to (91)
6.86%
No - I can't vote (<18, not in UK, etc) (632)
47.66%

Watched Threads

View All