perms and combs (1 Viewer)

[HeroWise]

i am getting 4/2454, can someone check

 

[HeroWise]

The answers did 4/11P4

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

 

[Drongoski]

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

 

[BenHowe]

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.

 

[HeroWise]

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]

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]

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

 

[HeroWise]

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]

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.