1. ## Permutations question help

Hi,

I have attached the questions below, can anyone help please?

2. ## Re: Permutations question help

1)a) Arrangements: 5! and 2 repetitions of 4, hence 5!/2! = 60

b) Not sure how to do the rest of the parts but I think just use cases. Hopefully someone like Integrand or Drongoski answers

2) There's only one scenario with CAL, and for the denominator, the ways to pick letters = 8C3, therefore 1/56

3. ## Re: Permutations question help

Ok for 1b since we want everything less 4000, need to consider all the outcomes for 3000. Note there should be 4 choose 3 = 4 different groupings of the 3 digits (after the intial digit) but since we want distinct answers there are only 3. If you write out all of them, there are 12. You then subtract this from a) to find the total >4000 i.e. 60-12=48 arrangements

c) For <5000 you will use part of b, you just need all the arrangements for the 4000's. Now with our four digits to choose from 3,4,5,6, they are all different so there will be 4 different groups i.e. 3,4,5 etc and the given arrangements of each group. If you do it all there are 24 different groups. So all the arrangements <5000= 24+12 =36

d) this is straightforward, from b) there are 12 possible arrangements <4000 and there are 60 total arrangements, let the outcome of an arrangement be A.

$P(A<4000)=\frac{12}{60}=\frac{1}{5}$

Mind you I couldve made an error lol

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