# perms and combs (1 Viewer)

#### HeroWise

##### Active Member

i am getting 4/2454, can someone check

Last edited:

#### HeroWise

##### Active Member
The answers did 4/11P4

I can understand what they did but arent they wrong. 11P4 means that M and M arent distinct

#### Drongoski

##### Well-Known Member
one way: Prob (CAME) = (1/11)x(2/10)x(2/9)x(1/8) = 1/1980

#### BenHowe

##### Active Member
The answers did 4/11P4

I can understand what they did but arent they wrong. 11P4 means that M and M arent distinct
No they arn't. probability=number of favourable outcomes/number of possible outcomes. This method is valid provided the number of possible outcomes is finite and equally likely.

Last edited:

#### HeroWise

##### Active Member
one way: Prob (CAME) = (1/11)x(2/10)x(2/9)x(1/8) = 1/1980

May I know why my way is wrong.

I understand how you got it, but why is my answer wayy off

2454 is the ways to arrange MATHEMATICS into 4 letters. Out of THEM CAME is part of them.

#### BenHowe

##### Active Member
May I know why my way is wrong.

I understand how you got it, but why is my answer wayy off

2454 is the ways to arrange MATHEMATICS into 4 letters. Out of THEM CAME is part of them.
Write out your method

#### BenHowe

##### Active Member
On a previous note 10P4 means M M -> M1,M2

#### HeroWise

##### Active Member
C1) 2 Alike 2 Alike
3C2* 4!/(2!2!)=18

C2) 2 alike 2 different
3C1 * 7C2* 4! /2! = 756

C3) all different
8C4 *4!=1680

Add that to get 2454

CAME is 2*2*1*1 /2454 = 4/2454

#### BenHowe

##### Active Member
Your listing or enumeration of cases is reasonable but it's not valid to use it as the possible outcomes since they arn't equally likely. You cover it at uni if you do combinatorics but just remember that for probability the outcomes must be equally likely.